Answer:
[tex]\large \boxed{\text{137 g CaH}_{2}}[/tex]
Explanation:
To solve this problem, we can use the Ideal Gas Law:
pV = nRT
1. Collect the data in one place:
M_r: 42.09
CaH₂ +2H₂O ⟶ Ca(OH)₂ +2H₂
p = 823 torr; V = 145 L; T = 21 °C
2. Convert temperature to kelvins
T = (21 +273.15) K = 294.15 K
3. Convert torrs to atmospheres.
[tex]p = \text{823 torr} \times \dfrac{\text{1 atm}}{\text{760 torr }} = \text{1.083 atm}[/tex]
4 Calculate the moles of H₂.
[tex]\begin{array}{rcl}\text{1.083 atm} \times \text{145 L} & = &n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{294.15 K}\\157.0 & = &24.14n \text{ mol}^{-1} \\n & = & \dfrac{157.0}{24.14 \text{ mol}^{-1} }\\\\ & = & \text{6.505 mol}\\\end{array}[/tex]
5. Calculate the moles of CaH₂
[tex]\text{Moles of CaH}_{2} = \text{6.505 mol H}_{2} \times \dfrac{\text{1 mol CaH}_{2}}{\text{2 mol H}_{2}} = \text{3.253 mol CaH}_{2}[/tex]
6. Calculate the mass of CaH₂
[tex]\text{Mass of CaH}_{2} = \text{3.253 mol CaH}_{2} \times \dfrac{\text{42.09 g CaH}_{2}}{\text{1 mol CaH}_{2}} = \text{137 g CaH}_{2}\\\text{The reaction needs $\large \boxed{\textbf{137 g CaH}_{\mathbf{2}}}$}[/tex]
