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14.7 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of...

14.7 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of carbon dioxide in atm is produced at 318 K in a 1.58 L flask. 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)

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Answer #1

Molar mass of C4H10 = 58.12 g/mol

mass of C4H10 = 14.7 g

mol of C4H10 = (mass)/(molar mass)

= 14.7/58.12

= 0.2529 mol

From balanced chemical reaction, we see that

when 2 mol of C4H10 reacts, 8 mol of CO2 is formed

mol of CO2 formed = (8/2)* moles of C4H10

= (8/2)*0.2529

= 1.0117 mol

we have:

V = 1.58 L

n = 1.0117 mol

T = 318.0 K

we have below equation to be used:

P * V = n*R*T

P * 1.58 L = 1.0117 mol* 0.08206 atm.L/mol.K * 318 K

P = 16.7 atm

Answer: 16.7 atm

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