## Module example question [ENDORSED]

curry 1E
Posts: 55
Joined: Wed Sep 18, 2019 12:15 am

### Module example question

18. For the following reaction if 1.00 x 102 g H2O reacts with 1.00 x 102 g CaC2 identify the limiting reactant.

CaC2(s) + H2O(l) ---> Ca(OH)2(aq) + C2H2(g)

I understand that you need to balance the equation first, then find the molar mass. But then what do you do after that?

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

### Re: Module example question  [ENDORSED]

After you balance the equation, you can write the stoichimetric ratios to make it easier to think about. After converting the 102 g of each substance into its respective number of moles, you take the reactants and using the molar ratios, determine how many moles of a particular product (just one of the products would be fine) it would produce. So for example, if the stoichimetric ratio from one of the reactants to products was 1:2, you would multiply the moles of reactants that you derived from the mass by 2 to arrive at moles of products. Do the same thing with the other reactant, and whichever one yields fewer moles of products is the limiting reactant. Hope that helps!

Kyle Thorin
Posts: 50
Joined: Sat Aug 24, 2019 12:16 am

### Re: Module example question

After calculating the molar mass of the reactants, you're going to want to calculate the number of moles in each reactant. 100g/(molar mass). You'll then get 1.56 moles of CaC2 and 5.55 moles of H2O. if you look at the balanced equation CaC2 + 2H2O -> Ca + C2H2, you'll see that one mole of CaC2 and 2 moles of H2O are required to make this equation work. So a 1 to 2 ratio is needed. Now, in order to tell which reactant is the limiting reactant, you must see which of the reactants get consumed the fastest. So, if you wanted to the reaction again, you would need 2 moles of CaC2 and 4 moles of H2O, but you can tell that you do not have enough CaC2, so CaC2 is the limiting reactant.