Question

gaseous butane (ch₃(ch₂)₂ch₃) reacts with gaseous oxygen gas (o₂) to produce gaseous carbon dioxide (co₂) and gaseous water (h₂o). if 1.48 g of carbon dioxide is produced from the reaction of 2.9 g of butane and 3.3 g of oxygen gas, calculate the percent yield of carbon dioxide. be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$$2CH_3(CH_2)_2CH_3 + 13O_2 ightarrow 8CO_2 + 10H_2O$$

Step2: Calculate the moles of butane and oxygen

Molar mass of butane \(C_4H_{10}\): \(M=(4\times12.01)+(10\times1.008)=58.12\space g/mol\)
Moles of butane \(n_{C_4H_{10}}=\frac{2.9\space g}{58.12\space g/mol}\approx0.05\space mol\)
Molar mass of \(O_2\): \(M = 32\space g/mol\)
Moles of \(O_2\) \(n_{O_2}=\frac{3.3\space g}{32\space g/mol}\approx0.103\space mol\)

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of \(C_4H_{10}\) to \(O_2\) is \(2:13\).
For \(0.05\space mol\) of \(C_4H_{10}\), moles of \(O_2\) required \(n = 0.05\times\frac{13}{2}=0.325\space mol\)
Since \(0.103\space mol\) of \(O_2\) is available (\(<0.325\space mol\)), \(O_2\) is the limiting reactant.

Step4: Calculate the theoretical moles of \(CO_2\)

From the balanced equation, mole ratio of \(O_2\) to \(CO_2\) is \(13:8\)
Moles of \(CO_2\) (theoretical) \(n_{CO_2}^{theo}=0.103\times\frac{8}{13}\approx0.0634\space mol\)

Step5: Calculate the theoretical mass of \(CO_2\)

Molar mass of \(CO_2\): \(M = 44\space g/mol\)
Theoretical mass \(m_{CO_2}^{theo}=0.0634\times44\space g/mol\approx2.79\space g\)

Step6: Calculate the percent yield

Percent yield \(=\frac{1.48\space g}{2.79\space g}\times100\%\approx53\%\)

Answer:

\(53\%\)