Question

c₃h₈(g) + 5o₂(g) → 3co₂(g) + 4h₂o(g)
knowing that o₂ is limiting, you want to
calculate the mass of c₃h₈ used up
when 105 l o₂ reacts at stp. what
mass of c₃h₈ is used up?
? g c₃h₈

Explanation:

Step1: Find moles of \( O_2 \) at STP

At STP, 1 mole of any gas occupies 22.4 L. So moles of \( O_2 \), \( n_{O_2}=\frac{V_{O_2}}{22.4\ L/mol}=\frac{105\ L}{22.4\ L/mol} \)

Step2: Molar ratio from reaction

From \( \ce{C3H8 + 5O2 -> 3CO2 + 4H2O} \), ratio of \( \ce{C3H8} \) to \( \ce{O2} \) is \( 1:5 \). So moles of \( \ce{C3H8} \), \( n_{C3H8}=\frac{1}{5}n_{O_2} \)

Step3: Calculate mass of \( \ce{C3H8} \)

Molar mass of \( \ce{C3H8} \) is \( 3\times12.01 + 8\times1.008 = 44.094\ g/mol \). Mass \( m = n_{C3H8}\times M \)

First, calculate \( n_{O_2}=\frac{105}{22.4}\approx4.6875\ mol \)

Then, \( n_{C3H8}=\frac{1}{5}\times4.6875 = 0.9375\ mol \)

Mass \( m = 0.9375\ mol\times44.094\ g/mol\approx41.3\ g \)

Answer:

\( 41.3 \) (or more precise calculation: \( \frac{105}{22.4\times5}\times44.094=\frac{105\times44.094}{112}\approx\frac{4630.87}{112}\approx41.35\approx41.4 \) depending on precision, but 41.3 is reasonable)