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 1.95 lb of each hydride at 750. torr and 27°c?

2758 L of LiH and 849 L of MgH₂

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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.