Question

During World War II, a portable source of hydrogen gas was needed for weather balloons, and solid metal hydrides were the most convenient form. Many metal hydrides react with water to generate the metal hydroxide and hydrogen. Two candidates were lithium hydride and magnesium hydride. What volume of gas is formed from 1.40 lb of each hydride at 750. torr and 27°C?

Answer 1

Answer:

1,981.84 L of gas is formed from 1.40 lb of lithium hydride.

1,268.36 L of gas is formed from 1.40 lb of magnesium hydride.

Explanation:

Hydrogen gas from lithium hydride:

$LiH+H_2O\rightarrow LiOH+H_2$

Mass of LiH = 1.40 lb = 635.0288 g

1 lb = 453.592 g

Moles of LiH = $\frac{635.0288 g}{8g/mol}=79.3786 mol$

According to reaction,1 mole of LiH gives 1 mole of hydrogen gas.

Then 79.3786 mole of LiH  will give :

$\frac{1}{1}\times 79.3786 mol=79.3786 mol$ of hydrogen gas.

Volume of hydrogen gas = V

Pressure of hydrogen gas , P= 750 Torr =0.987 atm

1 Torr = 0.00131579 atm

Temperature of the gas = T =  27°C = 300.15 K

Moles of gas = n = 79.3786 moles

$PV=nRT$

$V=\frac{nRT}{P}$

$=\frac{79.3786 mol\times 0.0821 atm L/mol K\times 300.15 K}{0.987 atm}=1,981.84 L$

Hydrogen gas from magnesium hydride:

$MgH_2+2H_2O\rightarrow Mg(OH)_2+2H_2$

Mass of $MgH_2$ = 1.40 lb = 635.0288 g

1 lb = 453.592 g

Moles of $MgH_2=\frac{635.0288 g}{25 g/mol}=25.4011 mol$

According to reaction,1 mole of $MgH_2$ gives 2 mole of hydrogen gas.

$\frac{2}{1}\times 26.4595 mol=50.8023 mol$ of hydrogen gas.

Volume of hydrogen gas = V

Pressure of hydrogen gas , P= 750 Torr =0.987 atm

1 Torr = 0.00131579 atm

Temperature of the gas = T =  27°C = 300.15 K

Moles of gas = n = 50.8023 moles

$PV=nRT$

$V=\frac{nRT}{P}$

$=\frac{50.8023 mol\times 0.0821 atm L/mol K\times 300.15 K}{0.987 atm}=1,268.36 L$