Answer:
46.4 L
Explanation:
We must first calculate the amount of H₂ and then calculate its volume.
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
[tex]M_r[/tex]: 42.09 18.02 2.016
[tex]\text{CaH}_{2} + \text{2H$_{2}$O} \longrightarrow \text{Ca(OH)}_{2}+\text{2H}_{2}[/tex]
Mass/g: 84.0 42.0
Step 2. Calculate the moles of each reactant
[tex]\text{Moles of CaH}_{2} = \text{84.0 g CaH}_{2} \times\frac{\text{1 mol CaH}_{2} }{\text{42.09 g CaH}_{2}} = \text{1.996 mol CaH}_{2}[/tex]
[tex]\text{Moles of H$_{2}$O}= \text{42.0 g H$_{2}$O} \times \frac{\text{1 mol H$_{2}$O}}{\text{2.016 g H$_{2}$O}} =\text{20.83 mol H$_{2}$O}[/tex]
Step 3. Identify the limiting reactant
Calculate the moles of H₂ we can obtain from each reactant.
From CaH₂:[tex]\text{Moles of H}_{2} = \text{1.996 mol CaH}_{2} \times \frac{\text{2 mol H}_{2} }{\text{1 mol CaH}_{2} } = \text{3.991 mol H}_{2}[/tex]
From O₂:[tex]\text{Moles of H}_{2} = \text{20.83 mol H$_{2}$O} \times \frac{\text{2 mol H}_{2} }{\text{2 mol H$_{2}$O} } = \text{20.83 mol H}_{2}[/tex]
The limiting reactant is CaH₂ because it gives the smaller amount of H₂.
Now, we use the Ideal Gas Law to calculate the volume.
pV = nRT
Divide both sides by p.
[tex]V = \frac{nRT }{p}[/tex]
[tex]p = \text{1463 torr} \times \frac{\text{1 atm}}{\text{760 torr}} = \text{ 1.925 atm}[/tex]
[tex]V = \frac{\text{3.991 mol}\times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1}\times \text{273 K} }{\text{ 1.925 atm}} = \text{46.4 L}[/tex]
