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Chapter 5: Problem 107

Lithium hydride reacts with water as follows: $$\mathrm{LiH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{LiOH}(a q)+\mathrm{H}_{2}(g)$$ 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 \(4,1-\mathrm{L}\) life belt at \(0.97 \mathrm{~atm}\) and \(12^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The mass of LiH required in grams can be calculated using the steps explained above.

Step by step solution

01

Calculate the number of moles of hydrogen gas

Using the ideal gas law, \(PV=nRT\), it's possible to solve for 'n', which is the number of moles of gas. Here, P = 0.97 atm, V = 4.1 L (converted from mL to L), R is the gas constant = 0.0821 L. atm/K. mol, and T is the absolute temperature in Kelvin (12 + 273.15 = 285.15 K). So, \(n = \frac{PV}{RT} = \frac{0.97 \times 4.1}{0.0821 \times 285.15}\)
02

Use Stoichiometry to Calculate Moles of LiH

In this reaction, each mole of LiH produces one mole of \(H_{2}\) gas. Therefore, the number of moles of \(H_{2}\) gas previously calculated is equal to the necessary number of moles of LiH.
03

Calculate the Mass of LiH in grams

Next, convert the moles of LiH to grams using its molar mass. The molar mass of LiH is 7.94 g/mol. Multiply the number of moles of LiH by its molar mass to get the mass in grams.

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