Question

chemistry 20
final exam 2025
gases

1. the volume of oxygen consumed at stp by combustion of 9.1 kg of propane

\\( \mathrm { c } _ { 3 } \mathrm { h } _ { 8 } ( \mathrm { g } ) \\) is
\\( \square 4.36 \times 10 ^ { 3 } \mathrm { l } \\) (the volume of a large closet)
\\( \square 2.31 \times 10 ^ { 4 } \mathrm { l } \\) (the volume of a small garage)
\\( \square 9.2 \times 10 ^ { 3 } \mathrm { l } \\) (the volume of a refrigerator)
\\( \square 4.60 \mathrm { l } \\) (the volume of a small propane tank)

Explanation:

Step1: Write the balanced chemical equation

The combustion of propane (\(C_3H_8\)) is \(C_3H_8+5O_2
ightarrow3CO_2 + 4H_2O\)

Step2: Calculate the molar mass of propane

The molar mass of \(C_3H_8\) is \(M=(3\times12)+(8\times1)=44\space g/mol\)

Step3: Convert the mass of propane to moles

Given mass \(m = 9.1\space kg=9100\space g\). Moles \(n=\frac{m}{M}=\frac{9100}{44}\space mol\approx206.82\space mol\)

Step4: Use the mole ratio from the balanced equation

From \(C_3H_8+5O_2
ightarrow3CO_2 + 4H_2O\), mole ratio of \(C_3H_8:O_2 = 1:5\). Moles of \(O_2\) consumed \(n_{O_2}=5\times206.82 = 1034.1\space mol\)

Step5: Use the ideal gas law at STP (\(T = 273\space K\), \(P=1\space atm\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\))

\(V=\frac{nRT}{P}\). Substituting \(n = 1034.1\space mol\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 273\space K\), \(P = 1\space atm\)
\(V=1034.1\times0.0821\times273\)
\(V = 1034.1\times22.4\) (since \(0.0821\times273\approx22.4\) at STP for molar volume)
\(V\approx2.31\times10^{4}\space L\)

Answer:

2.31 × 10⁴ L (the volume of a small garage)