Calcium hydride (CaH₂, 42.10 g/mol) is so reactive with water 18.02 g/mol) that it can be used to remove traces of water from nonaqueous solvents.What mass of the excess reactant will be left over if 24.6 g CaH₂ is added to a large volume of solvent that contains 14.0 g H₂O?
CaH₂ (s) +2H₂ O(l) $\rightarrow$ Ca(OH) ₂ (s) + 2H₂ (g)

Question

Calcium hydride (CaH₂, 42.10 g/mol) is so reactive with water 18.02 g/mol) that it can be used to remove traces of water from nonaqueous solvents.What mass of the excess reactant will be left over if 24.6 g CaH₂ is added to a large volume of solvent that contains 14.0 g H₂O?
CaH₂ (s) +2H₂ O(l)

\rightarrow

Ca(OH) ₂ (s) + 2H₂ (g)

Quizplus · Accepted Answer

First, calculate the moles of each reactant: $ \text{moles of CaH}_2 = \frac{24.6 \, \text{g}}{42.10 \, \text{g/mol}} $ and $ \text{moles of H}_2\text{O} = \frac{14.0 \, \text{g}}{18.02 \, \text{g/mol}} $. The reaction stoichiometry shows that 1 mole of $ \text{CaH}_2 $ reacts with 2 moles of $ \text{H}_2\text{O} $, so the limiting reactant can be determined by comparing the actual mole ratio to the stoichiometric ratio. The calculation shows that $ \text{H}_2\text{O} $ is the limiting reactant, and there will be excess $ \text{CaH}_2 $ left over. Calculating the exact amount of $ \text{CaH}_2 $ left requires determining how much reacts and subtracting this from the initial amount.

Calcium Hydride (CaH2, 42 →\rightarrow→ Ca(OH)2 (S) + 2H2 (G) A)8

Calcium Hydride (CaH₂, 42 $\rightarrow$ Ca(OH)₂ (S) + 2H₂ (G)
A)8