What mass of excess reagent remains when 4.70 g of SrH2 and 3.75 g of H2O react according to the following reaction? The percentage yield is 100%. SrH2(s) + 2 H2O(l) → Sr(OH)2(s) + 2 H2(g)

Question

Molar masses (in g/mol): SrH₂=89.64, H₂O=18.02, Sr(OH)₂=121.64, H₂=2.02

How would I solve this? And what compound would be the excess reagent? H₂O or Sr(OH)₂?

1. A) 0.95 g
2. B) 1.86 g
3. C) 6.27 g
4. D) 6.38 g
5. E) 12.7 g

Answer 1

SrH2(s) + 2 H2O(l) → Sr(OH)2(s) + 2 H2(g) ... balanced equation

First, find which reactant is limiting. We can do this by finding mass of Sr(OH)2 or mass of H2 produced. For simplicity, I'll find mass of H2 produced from each reactant.

For SrH2: 4.70 g SrH₂ x 1 mol/89.64 g x 2 mol H₂/mol SrH₂ x 2.02g H₂/mol = 0.2118 g H₂ produced

For H2O: 3.75 g H₂O x 1 mol/18.02 g x 2 mol H₂/2 mol H₂O x 2.02 g H₂/mol = 0.4204 g H₂ produced

Therefore SrH₂ would be limiting as it produces fewer grams of H₂

Since SrH₂ is limiting, H₂O must be in excess. To find out how much H₂O is left over, we find out how much H₂O was used up when it reacted with all of the SrH₂, i.e when it reacted with 0.4.70 g of SrH₂:

moles SrH₂ used up = 4.70 g SrH₂ x 1 mol/89.64 g = 0.0524 moles SrH₂ used up (all of it)

moles H₂O used up = 0.0524 mol SrH₂ x 2 mol H₂O/1 mol SrH₂ = 0.1048 moles H₂O used up

moles H₂O originally present = 3.75 g H₂O x 1mol H2O/18.02 g = 0.2081 moles H₂O originally present

moles of H₂O left over after reaction = 0.2081 mol - 0.1048 mol = 0.1033 moles H₂O left over

mass H₂O left over = 0.1033 moles x 18.02 g/mole = 1.86 g H₂O left over.