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 3.6 L life belt with hydrogen gas at 1.04 atm and 27.7°C?

Follow the steps of this worked example.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To answer this question, we need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the given temperature from Celsius to Kelvin.

T = 27.7°C + 273.15 = 300.85 K

Next, let's convert the given pressure from atm to kPa (since the ideal gas constant, R, has units of kPa⋅L/mol⋅K).

P = 1.04 atm * 101.325 kPa/atm = 105.38 kPa

Now, we can rearrange the ideal gas law equation to solve for the number of moles of hydrogen gas (n):

n = PV / RT

n = (105.38 kPa * 3.6 L) / (8.314 kPa⋅L/mol⋅K * 300.85 K)

Simplifying the units, we get:

n = (105.38 kPa * 3.6 L) / (8.314 kPa⋅L/K⋅mol * 300.85 K)
n = 0.015 mol

According to the balanced chemical equation, 1 mole of LiH reacts to produce 1 mole of H2 gas. Therefore, we need 0.015 mol of LiH to produce the required amount of H2 gas.

The molar mass of LiH is approximately 7 + 1 = 8 g/mol. Therefore, we can calculate the mass of LiH required:

mass = moles * molar mass
mass = 0.015 mol * 8 g/mol
mass = 0.12 g

Hence, approximately 0.12 grams of LiH are needed to fill a 3.6 L life belt with hydrogen gas at 1.04 atm and 27.7°C.