[Music] good day everyone and this is a continuation of our lesson in reaction stoichiometry so these are just special topics involving stoichiometry of solutions and gases so we will just be solving some additional examples for stoichiometry of solutions and gases so let's begin okay let's start with solution stoichiometry solution stoichiometry is quite commonly encountered know especially in your laboratory so for this we have to recall the concept of molarity molarity is just one of the few ways on how we can express the concentration of a solute in a solution molarity is defined as the number of moles of your solute divided by the volume of the solution in liters okay normally in these types of problems your solvent is almost always going to be water for it to become an aqs system let's apply this concept of molarity to this example a 2.5 gram piece of magnesium is dropped into a beaker containing 50 ml of 1.20 molar hydrochloric acid or hcl and this is our reaction magnesium plus hcl yields mgcl2 plus hydrogen we are to determine how much magnesium chloride in grams is obtained from the reaction so let's first write the given our reaction is magnesium plus hcl yields magnesium chloride and hydrogen gas let's take a look at the numerical given so we have 2.5 grams of magnesium and for hcl you have two parameters given here the volume of the solution and the molarity of the solution both of these numbers are very important because you want to determine the number of moles of your solute in this case that is hcl so we write that we have 50 ml of 1.20 molar hcl it is from these two numbers that we determine the number of moles of hcl okay and the problem is asking us for the mass of magnesium chloride that was produced from this reaction okay let's start with the hcl again the definition of molarity is moles of solute in this case your solute is hcl divided by the volume of the solution in liters we are given the molarity you are also given the volume of the solution we just have to convert it to liters and then we can solve for the number of moles of hcl so by cross multiplication we get that the number of moles of hcl is equal to the molarity multiplied by the volume of the solution so let's compute the molarity is 1.2 molar and the volume of the solution is 50 ml although we have to convert this to liters you simply divide by 1000 ml is to 1 liter okay so the number of moles of hcl is 1.2 times 50 divided by 1000 that is 0.06 moles hcl the next step is to determine which among magnesium and hcl is your limiting reactant again it's important that you do not forget your concept of the limiting reactant because even though it's not being stated explicitly in the problem you still have to consider solving for the limiting reactants okay so let me first just clear some of this we now know that the amount of hcl is 0.06 moles now let's convert the amount of magnesium to moles and then let's see how much magnesium chloride can we produce so we start with 2.5 grams of magnesium let's take a look at the molecular weight of magnesium magnesium is here that's 24.305 grams per mole okay and then let's take a look at the balance equation by the way we have forgotten to balance the equation so balancing our equation you take a look at magnesium so it's already balanced you have one magnesium on both sides hydrogen you have two hydrogen in the product and one on the reactant so we have to multiply hcl by two and consequently that also balances chlorine two chlorine on the reactant to chlorine on the product we're good to go okay based on the balanced equation if we're solving for the amount of magnesium chloride for every one mole of magnesium there is one mole of magnesium chloride and since the problem is already asking us for the amount of magnesium chloride in grams let's continue our conversion to grams we simply want to convert the number of moles of foundation chloride to grams of magnesium chloride and for that we need to get the molecular weight of magnesium chloride let's go back to our periodic table and let's take a look at magnesium and chlorine magnesium has a molecular rate of 24.305 and chlorine has a molecular weight of 35.45 so that would be 24.305 for magnesium plus 35.45 times 2 for chlorine the molecular weight is 95.205 grams per mole we need to multiply the number of moles by the molecular weight so this becomes 95.205 grams of magnesium chloride for every one mole of magnesium chloride you know that we are on the right track because when you cancel the units grams of magnesium cancels mole of magnesium cancels and mole of magnesium chloride cancels your remaining unit is grams of magnesium chloride okay let's solve for that that would be 2.5 divided by 24.305 times 95.205 we have 9.79 grams of magnesium chloride however this is not yet our final answer okay we still have to take a look at how much magnesium chloride will be produced coming from hcl so if you start from hcl you have 0.06 moles of hcl it is already in moles so we can now go to the balanced equation the balance equation states that there are two moles of hcl for every one mole of magnesium chloride and then we can convert the moles of fungus chloride to mass by using the same factor that we used earlier the molecular weight we multiply this by 95.205 grams of magnesium chloride per mole of magnesium chloride cancel the units moles you're left with grams of magnesium chloride so that is .06 divided by 2 times 95.205 and our answer is 2.86 grams of magnesium chloride based on our two answers what do we choose we choose the lower one and we say that hcl is our limiting reactant and magnesium is our excess reactants so we say that the 9.79 grams of magnesium chloride is the wrong answer because it came from the excess reactants going back to the problem we are just asked to determine how much magnesium chloride is obtained from the reaction our answer is 2.86 grams okay that is how we apply stoichiometry in solutions you just have to remember the concept of molarity let's have another example calcium carbonate is known to be insoluble in water if you mix 50 ml of a 0.125 molar calcium chloride solution with 50 ml of a 0.112 molar sodium carbonate solution what mass of calcium carbonate can be recovered this is another one of those examples wherein you're not given the chemical equation so we have to decipher the chemical equation from the given based on the problem we are mixing two solutions one sodium carbonate and the other is calcium chloride so let's see what happens when we mix those two sodium carbonate plus calcium chloride this looks like a double displacement reaction wherein the sodium will be partnered with chlorine and then the calcium will be parted with the carbonate because the problem asks us how much calcium carbonate can be recovered okay so our products in this case would be sodium chloride plus calcium carbonate based on the double displacement reaction now to balance the equation we simply multiply sodium chloride by two to check sodium is balanced that's two is to two chlorine is also balanced two is to two carbon one is to one and oxygen that's three in the reactant and three in the product so this is already a balanced equation now let's take a look at the other given you're given the volume and the concentration of both calcium chloride and sodium carbonate so for calcium chloride that is 50 ml of a 0.125 molar solution well for sodium carbonate that's still 50 ml but the concentration is now 0.112 molar and we are solving for the mass of calcium carbonate okay so our given is now complete now you just have to recall again the concept of molarity again molarity is equal to the moles of the solute divided by the volume of the solution in liters we are already given the volume we're also given the concentration in molar for both of our reactants so we simply have to convert them to moles and then determine which among them is the limiting reactant then we can solve for the amount of calcium carbonate that can be formed okay so based on our formula here if you want to determine the moles of the solute you simply have to get the products of the concentration in molarity and the volume of the solution in liters so let's do that for sodium carbonate we simply multiply 50 ml but divide it with 1000 to make it litters remember that the denominator of molarity is always liters so if you're given a solution volume in milliliters do not forget to convert it to liters okay so that is 50 divided by 1000 to convert it to liters multiplied by the molarity of sodium carbonate solution 0.112 the number of moles of sodium carbonate is 5.6 times 10 raised to negative 3 moles next we do the same for calcium chloride that is 50 divided by 1000 to convert it to liters multiplied by the concentration 0.125 the answer is 6.25 times 10 raised to negative 3 moles of calcium chloride now let's determine which among the 2 is the limiting reactants and let's determine the amount of calcium carbonate that can be formed let's start with sodium carbonate that's 5.6 times 10 raised to negative 3 moles of sodium carbonate this is already in moles so we can proceed with the balanced equation for every one mole of sodium carbonate we have one mole of calcium carbonate as well and we want to convert the moles of calcium carbonate to mass because we are asked for the mass of calcium carbonate so let's solve for its molecular weight the molecular weight of calcium is 40.078 so that is 40.078 for calcium plus for carbon that is 12.011 and then three oxygen at 15.999 each the molecular weight of calcium carbonate is 100.086 grams of calcium carbonate for every mole of calcium carbonate cancel the units what do we have that is 5.6 times 10 raised to negative 3 times 100.086 we get 0.56 grams of calcium carbonate if we started with sodium carbonate but at this point we have nothing to compare it yet so we need to solve the one from calcium chloride we start with 6.25 times 10 raised to negative 3 moles of calcium chloride and then based on the balanced equation for every 1 mole of calcium chloride there's also 1 mole of calcium carbonate and then we use again the molecular weight of calcium carbonate 100.086 grams of calcium carbonate per mole we can determine the mass of calcium carbonate produced that is 6.25 times 10 to the negative 3 times 100.086 we have 0.626 so which among the two values do we choose we choose the lesser value we say that we can generate 0.56 grams of calcium carbonate and that sodium carbonate is our limiting reactants okay we cannot generate 0.626 grams of calcium carbonate because that would be in violation of the limiting reactants okay so it's basically the same approach in stoichiometry but we are now adding the definition of the molarity okay so always keep this in mind let's continue for the final part of this additional lecture we have stoichiometry involving gases the stoichiometry aspect would still be the same however the main difference here is that we are dealing with gases so we have to use the ideal gas law because the number of moles is very hard to measure in gases we often rely on other parameters such as temperature pressure and volume to determine the amount of gases present okay if you will recall the ideal gas law is characterized by this equation that's pv equals nrt where in p is pressure v is volume n is the number of moles of the gas r is the real gas constant that's 0.08206 liter atmosphere per mole kelvin and p is the temperature so you only need to know the pressure the volume and the temperature to know the number of moles of the gas and it is this number of moles of the gas that we can use in stoichiometry let's try to solve this example we basically have the same problem a 2.50 gram piece of magnesium is dropped into a beaker containing 50 ml of 1.20 molar of hcl it's the same reaction and the same given however this time we want to determine how many liters of hydrogen at 25 degrees celsius and one atmosphere is obtained from the reaction so we are now interested in the amount of hydrogen here's our given so this given is the same as that of our previous example involving the same reaction from that problem we have determined that the number of moles of hcl is 0.06 moles and that hcl is our limiting reactants but in this case we want to determine what volume of hydrogen is being produced given that the temperature is 25 degrees celsius and the pressure is one atmosphere let's try to solve this since we know what the limiting reactant is we can already proceed to the determination of the number of moles of our products which in this case is hydrogen so we start with 0.06 moles of our limiting reactant hcl and we want to determine how many moles of hydrogen can we produce based on the balanced equation for every two moles of hcl we produce one mole of hydrogen gas so our answer is 0.03 moles of hydrogen gas and with the number of moles of hydrogen gas given the temperature and pressure we can easily solve for the volume of the gas using the ideal gas law the ideal gas law pv equals nrt we want to determine the volume so we divide both sides by the pressure we get nrt over p and we can substitute everything and solve for the volume solving for the volume the number of moles as we have determined is 0.03 moles of hydrogen gas multiplied by r remember that r is a constant that's 0.08206 liter atmosphere per mole per kelvin and then temperature is always in kelvin so you need to convert 25 degree celsius to kelvin you simply add 273.15 to make it kelvin divided by the pressure of one atmosphere now if you want to double check if you are correct in the derivation of the formula coming from the ideal gas law you can try canceling the units moles would cancel kelvin would cancel and atmosphere will also kinda sell that leaves you with only one unit and that is liters so we know that we are correct okay solving we can solve for the volume of the hcl that would be 0.03 times 0.08206 times 25 plus 273. fifteen divided by one that is zero point seven three four liters okay that means that we can produce 0.734 liters of hydrogen coming from this reaction at 25 degrees celsius and one atmosphere okay the approach is still the same the only different thing in this case is that this involves gases and we have used the ideal gas law for the computation of volume otherwise it's the same process let's solve for the last example kalburo or calcium carbide is a compound used by fruit growers to artificially ripen fruits such as mangoes and bananas when reacted with water it produces acetylene and calcium hydroxide according to this reaction that's calcium carbide plus water yields acetylene plus calcium hydroxide if 55 grams of kalburo or calcium carbide is reacted with a large excess of water that are mean the volume of acetylene that can be recovered at 30 degrees celsius and atmospheric pressure okay so let's list the given you are given here the complete chemical reaction so that's good and it was also stated that water is provided in excess that means that water is your excess reactant and kalburo or calcium carbide is your limiting reactants so we are given 55 grams of kalburo we want to determine the volume of acetylene at 30 degrees celsius and atmospheric pressure and here is our given let's first try to balance our equation calcium is already balanced carbon 2 is the two that's already balanced hydrogen we have two in the reactant and you have four in the product so you have to multiply water by 2 to make it balanced and oxygen is 2 in the reactant and 2 also in the products okay so what is our plan here our general plan is to take the mass of the calcium carbide convert it to moles of calcium carbide and then use the balance equation to turn it to the number of moles of acetylene and then later use the ideal gas law to convert from moles of acetylene gas to the liters of acetylene okay let's begin we have 55 grams of calcium carbide we need the molecular weight of calcium carbide calcium is 40.0 plus for carbon that is 12.011 times 2. the molecular weight is 64.1 grams of calcium carbide is 2 moles of calcium carbide next we take a look at the balance equation and and look at the coefficients for every one mole of calcium carbide there is one mole of acetylene or c2h2 okay now we can solve for the number of moles of calcium carbide that is 55 divided by 64.1 that's 0.858 moles c2 h2 now as with our previous problem at this point we simply need the ideal gas law let's pv equals nrt solving for volume that becomes nrt over p and substitute everything we can find the volume of acetylene okay so the volume is equal to number of moles 0.858 and then r 0.08206 liter atmosphere per mole kelvin multiplied by the temperature 30 degrees celsius converting to kelvin plus 273.15 divided by the pressure one atmosphere solving is point eight five eight times point zero eight two zero six times thirty plus two seventy three point fifteen we get twenty one 21.3 liters of acetylene okay so you see that the solution is just repetitive you just have to master how to perform stoichiometry and then apply additional equations such as the ideal gas equations for stoichiometry involving gases and the definition of molarity for solution stoichiometry please watch out for another video release detailing the solutions for more examples so that you can practice okay that's the end of this video i hope you have learned something if you have questions you can always send me a message keep practicing and as always keep safe [Music] you