Determine the volume of air required for combustion of 10 liters of pentane.

univerkov | March 13, 2021 | education

Let’s find the amount of the substance pentane C5H12.

n = V: Vn, where Vn is the molar volume of gases equal to 22.4 l / mol.

n = 10 L: 22.4 L / mol = 0.446 mol.

C5H12 + 8O2 = 5CO2 + 6H2O.

For 1 mol of С5Н12, there is 8 mol of О2.

n (O2) = 8n (C5H12) = 0.446 × 8 = 3.5 68 mol.

Let’s find the volume of O2.

V = n Vn, where Vn is the molar volume of gas equal to 22.4 l / mol.

V = 3.568 mol × 22.4 L / mol = 79.92 L.

79.92 L – 21%,

V l – 100%,

V = (79.92 l × 100%): 21% = 380 l.

Answer: 380 l.

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