Question

If 6.5 g of zinc reacts with 5.0 g of HCl, according to the following reaction.
Zn + 2HCl → ZnCl2 + H2
How many grams of the reactant remain unreacted?
c. How many grams of hydrogen would be produced?

Answer 1

Zn + 2HCl → ZnCl2 + H2
6.5 g Zn and 5.0 g HCl. This is a limiting reagent (LR) problem. You know that when an amount for more than one of the reactants is given. First determine the LR.
mols Zn = g/atomic mass = 6.5/65.4 = 0.0994
mols HCl = 5.0/36.5 = 0.137
If Zn is the LR, how much HCl is needed? That's 0.0994 x (2 mol HCl/1 mol Zn) = 0.0994 x 2 = 0.199 needed. That much HCl is not present (only 0.137 mols HCl present); therefore, HCl must be the LR and Zn is the excess reagent (ER).
c. Using the HCl will produce how many mols H2? That's 0.137 x (1 mol H2/2 mols HCl) = 0.137/2 = 0.0685. Convert that to grams; i.e., grams H2 = mols H2 x molar mass H2 = ?
b. Zn is the ER. So how much of the Zn will be used up when all of the HCl is gone. That's 0.137 mols HCl x (1 mol Zn/2 mols HCl) = 0.137/2 = 0.0685. Convert that to grams. 0.0685 mols Zn x 65.4 g/mol = 4.48 g Zn used. How much remains? That's 5.0 starting - 4.48 g used = ? grams ER remaining.
Post your work if you get stuck.