Respuesta :
Answer: The mass of [tex]Ni(CO)_4[/tex] allowable in the laboratory is 4599.5 grams
Explanation:
To calculate the volume of cuboid, we use the equation:
[tex]V=l\times b\times h[/tex]
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:
[tex]V=14\times 22\times 9=2772ft^3=78503.04L[/tex] (Conversion factor: [tex]1ft^3=28.32L[/tex]
To calculate the moles of gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 1.00 atm
V = Volume of the gas = 78503.04 L
T = Temperature of the gas = [tex]23^oC=[23+273]K=296K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]1.00atm\times 78503.04L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 296K\\\\n=\frac{1.00\times 78503.04}{0.0821\times 296}=3230.4mol[/tex]
Applying unitary method:
For every 109 moles of gas, the moles of [tex]Ni(CO)_4[/tex] present are 1 moles
So, for 3230.4 moles of gas, the moles of [tex]Ni(CO)_4[/tex] present will be = [tex]\frac{1}{109}\times 3230.4=26.94mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of [tex]Ni(CO)_4[/tex] = 26.94 moles
Molar mass of [tex]Ni(CO)_4[/tex] = 170.73 g/mol
Putting values in above equation, we get:
[tex]26.94mol=\frac{\text{Mass of }Ni(CO)_4}{170.73g/mol}\\\\\text{Mass of }Ni(CO)_4=(26.94mol\times 170.73g/mol)=4599.5g[/tex]
Hence, the mass of [tex]Ni(CO)_4[/tex] allowable in the laboratory is 4599.5 grams
