Question

Lithium hydride reacts with water as follows.
LiH(s) + H2O(l) LiOH(aq) + H2(g)

During World War II, U.S. pilots carried LiH tablets. In the event of a crash landing at sea, the LiH would react with the seawater and fill their life belts and lifeboats with hydrogen gas. How many grams of LiH are needed to fill a 4.1 L life belt at 0.97 atm and 28°C?

Answer 1

Answer: 1.4 g LiH

Explanation:

In order to solve this problem, we must use the ideal gas law(shown below).

$\begin{document}\\Ideal~ Gas ~Law:\[\boxed{PV = nRT}\]\\\\Where:\\P = \text{pressure (in atm)} \\V = \text{volume (in liters)} \\n = \text{number of moles of gas} \\R = \text{ideal gas constant (\(0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\))} \\T = \text{temperature (in Kelvin)}\end{document}$

In this problem:

$Given:\\P = \boxed{0.97 \,\text{atm} }\\V = \boxed{4.1 \,\text{L}} \\T = \boxed{28^\circ C = 301 \,\text{K}} \\\text{Molar mass of } LiH = \boxed{7.95 \, \text{g/mol}}\end{align*}$

Now plug into the ideal gas law(for the above we needed to convert Celsius to Kelvin by adding 273 to it):

Solving:

$\[n = \frac{PV}{RT}\\\[n = \frac{(0.97 \, \text{atm})(4.1 \, \text{L})}{(0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}})(301.15 \, \text{K})}\]\[n \approx \frac{(3.977)}{(24.744)} \approx \boxed{0.170 \, \text{moles of} \, H_2}\] Convert moles of hydrogen gas to grams of lithium hydride(Same molar value:)\[\text{Mass} = n \times \text{Molar mass}\]\[\text{Mass} = 0.170 \, \text{moles} \times 7.95 \, \text{g/mol}\]$

$\[\text{Mass} \approx \boxed{\boxed{1.4 \, \text{grams} \, LiH}}\]$

Since the question gives us only 2 significant figures, our final answer will be 1.4 grams LiH.