n the reaction, Zn + 2 HCl  ZnCl2 + H2, 25.0 grams of Zn are reacted with 17.5 g of HCl. How many grams of H2 will be produced?

Answer 1

From the balanced equation and the molar masses of Zn, HCl, and H2, we can determine the theoretical yield of H2. This is a "limiting reactant" question (because we are given the masses of both reactants & we don't know which is in excess & which is limiting). Hence we need to do two separate calculation. The one producing the lower mass of H2 gives us the answer.

Let's first consider using the Zn:

(2.02 g H₂/1 mol H₂) (1 mol H₂/1 mol Zn) (1 mol Zn/65.38 g Zn) 25.0 g Zn = 0.7724 = 0.772 g H₂

So, if the Zn is limiting then we get a yield of 0.772 g H₂

Now, if the HCl is limiting:

(2.02 g H₂/1 mol H₂) (1 mo H₂/2 mol HCl) (1 mol HCl/36.46 g HCl) 17.5 g HCl = 0.484 g H₂

The HCl produces the lower amount of H₂, so the mass of hydrogen produced is 0.484 g. The HCl is the limiting reactant & the Zn is in excess.