A drum used to transport crude oil has a volume of 162 L. How many grams of water, as steam, are required to fill the drum at 1.20 atm and 100°C?

114.48g

Explanation:

Step 1:

Data obtained from the question:

Volume (V) = 162 L

Pressure (P) = 1.20 atm

Temperature = 100°C = 100°C+ 273 = 373K

Number of mole (n) of H2O =?

Recall:

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the number of mole of H2O i.e steam.

Using the ideal gas equation, the number of mole of H2O can be obtained as follow:

PV = nRT

1.2x 162 = n x 0.082 x 373

Divide both side by 0.082 x 373

n = (1.2x 162) /(0.082 x 373)

n = 6.36 moles

Step 3:

Determination of mass of H2O i.e steam:

This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole H2O = 6.36 moles

Mass of H2O =?

Mass = number of mole x molar Mass

Mass of H2O = 6.36 x 18

Mass of H2O = 114.48g

Therefore, the mass of H2O i.e steam required to fill the drum of crude oil is 114.48g