<u>Answer:</u> The mass of
allowable in the laboratory is 4599.5 grams
<u>Explanation:</u>
To calculate the volume of cuboid, we use the equation:

where,
V = volume of cuboid
l = length of cuboid = 14 ft
b = breadth of cuboid = 22 ft
h = height of cuboid = 9 ft
Putting values in above equation, we get:
(Conversion factor: 
To calculate the moles of gas, we use the equation given by ideal gas which follows:

where,
P = pressure of the gas = 1.00 atm
V = Volume of the gas = 78503.04 L
T = Temperature of the gas = ![23^oC=[23+273]K=296K](https://tex.z-dn.net/?f=23%5EoC%3D%5B23%2B273%5DK%3D296K)
R = Gas constant = 
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:

Applying unitary method:
For every 109 moles of gas, the moles of
present are 1 moles
So, for 3230.4 moles of gas, the moles of
present will be =
To calculate the number of moles, we use the equation:

Moles of
= 26.94 moles
Molar mass of
= 170.73 g/mol
Putting values in above equation, we get:

Hence, the mass of
allowable in the laboratory is 4599.5 grams