in today's video we're going to look at limiting reactants and see how they affect our calculations when we try to work out the mass of a product to see what a limiting reactant is let's imagine that we had a beaker of hydrochloric acid and we dropped in some calcium carbonate as the hydrochloric acid and calcium carbonate react they'll produce calcium chloride water and carbon dioxide we will be able to see that a reaction is taking place because the solid calcium carbonate will slowly disappear and bubbles of gas will be given off which is our carbon dioxide after a while though all the calcium carbonate will disappear and the fizzing will slow down and eventually stop so at this point we know that one of our reactants must have all been used up because the reaction stopped if we can see that the calcium carbonate has all disappeared then we know that it must have all been used up and so we would call it the limiting reactant because it limited how much product could be produced on the other hand there's probably still loads of hydrochloric acid left so we'd say that the hydrochloric acid is in excess the reason this is important is because how much product we get depends entirely on the limiting reactant for example if we add more calcium carbonate we'll see more fizzing as more products are made however this fizzing would stop as the calcium carbonate is used up again whereas if we add some more hydrochloric acid then nothing will happen because it has no calcium carbonate to react with now when it comes to calculations you'll sometimes be told that one of the reactants is in excess in other cases though it might just be implied in the question for example calculate the mass of sodium oxide produced when 115 grams of sodium is burned in air burning in air is known as combustion and basically just means reacting with oxygen and if you think about it there's loads of oxygen floating around in the air so we can assume that the oxygen will be the one in excess which means that our sodium must be the limiting reactant the first step with any question like this is to write out the equation and then balance it and if you're not comfortable balancing equations i'll link our video on it below in the description as we want the mass of sodium oxide we're first of all going to have to figure out the number of moles of sodium oxide that are going to be produced and to do that we have to work out how many moles of sodium we started with for this we use the equation moles equals mass over mr so 115 grams of sodium divided by its mr which is 23 giving us 5 moles of sodium then by looking at our balanced equation we can find the ratio between the sodium and sodium oxide which is four to two or more simply two to one and so if we use five moles of sodium we know we must produce 2.5 moles of sodium oxide so now we need to find what the mass of two and a half moles of sodium oxide is to do this we're gonna have to take our equation and rearrange it to get mass equals moles times mr we already know the number of moles is 2.5 so we just need to get the mr of sodium oxide which we get by taking 2 times 23 because there are two sodiums in na2o and adding 16 for the oxygen giving us 62. then we can plug these in to get 2.5 times 62 giving us 155 grams which is the mass of sodium oxide that we'll get anyway that's all for this video so hope you found it useful and we'll see you next time you