Using Limiting Reagents
So what? What's the point of identifying the limiting reagent in a chemical reaction. Well take the s'mores example: If you decided to buy one package of graham crackers and more chocolate then that would be your limiting reagent. Now that you know this you'd want to know how many s'mores you could make in total. In this case that would be as many s'mores as you could make using one package of graham crackers as the limiting reagent which would be 10 because there are 20 graham crackers in a package and it takes two graham crackers to make a s'more. But in chemical equations it's not so simple.
Take this case:
For the balanced equation shown below, if 46.0 grams of CaH2 were reacted with 58.0 grams of H2O, how many grams of Ca(OH)2 would be produced?
CaH2+2H2O=>Ca(OH)2+2H2
So in this case you would find the limiting reagent (for a reminder on how to do this click on this link)
The limiting reagent in this equation is: CaH2 is the limiting reagent and there are 1.095 moles of it.
Now to find out how many grams of Ca(OH)2 would be produced you need to find out how many moles of it are produced, to do this you find the ratio of each coefficient in this case that is 1:1, and then you maintain that same ratio to find the number of moles produced. 1/1=1.095 moles/x
So you have 1.095 mol of Ca(OH)2 which you then can use to figure out grams by multiplying it by the gfw of Ca(OH)2 which is 74. This equals 81.03 grams
So 81.03 grams of Ca(OH)2 are produced.
Another example (with a harder ratio)
For the balanced equation shown below, if 10.9 grams of CH2Br2 were reacted with 1.87 grams of O2, how many grams of CO2 would be produced?
2CH2Br2+3O2=>2CO2+2H2O+2Br2
The limiting reagent is O2 which there are 0.05 moles of. The ratio of the coefficients is 3:2
3/2= 0.05 mol/x mol
x=0.03 mol,
0.33 mol X 44(gfw of CO2) =1.4666 grams
Take this case:
For the balanced equation shown below, if 46.0 grams of CaH2 were reacted with 58.0 grams of H2O, how many grams of Ca(OH)2 would be produced?
CaH2+2H2O=>Ca(OH)2+2H2
So in this case you would find the limiting reagent (for a reminder on how to do this click on this link)
The limiting reagent in this equation is: CaH2 is the limiting reagent and there are 1.095 moles of it.
Now to find out how many grams of Ca(OH)2 would be produced you need to find out how many moles of it are produced, to do this you find the ratio of each coefficient in this case that is 1:1, and then you maintain that same ratio to find the number of moles produced. 1/1=1.095 moles/x
So you have 1.095 mol of Ca(OH)2 which you then can use to figure out grams by multiplying it by the gfw of Ca(OH)2 which is 74. This equals 81.03 grams
So 81.03 grams of Ca(OH)2 are produced.
Another example (with a harder ratio)
For the balanced equation shown below, if 10.9 grams of CH2Br2 were reacted with 1.87 grams of O2, how many grams of CO2 would be produced?
2CH2Br2+3O2=>2CO2+2H2O+2Br2
The limiting reagent is O2 which there are 0.05 moles of. The ratio of the coefficients is 3:2
3/2= 0.05 mol/x mol
x=0.03 mol,
0.33 mol X 44(gfw of CO2) =1.4666 grams