Answer:
2758 L of LiH and 849 L of MgH₂
Step-by-step explanation:
To know the volume of the gas, we can use the ideal gas law:
PV =nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (62.36 torr*L/mol*K), and T is the temperature.
The mass is 1.95 lb = 884.52 g (1lb = 453.6 g).
T = 27ºC + 273 = 300 K
For lithium hydride (LiH)
The molar mass is: 7 g/mol of Li + 1 g/mol of H = 8g/mol
n = 884.52/8 = 110.565 mol
750*V = 110.565*62.36 *300
V = 2758 L of LiH
From magnesium hydride (MgH₂)
The molar mass is: 24 g/mol of Mg + 2* 1 g/mol of H = 26 g/mol
n = 884.52/26 = 34.02 mol
750*V = 34.02*62.36 *300
V = 849 L of MgH₂