Question

gaseous butane (ch₃(ch₂)₂ch₃) reacts with gaseous oxygen gas (o₂) to produce gaseous carbon dioxide (co₂) and gaseous water (h₂o). if 90.2 g of carbon dioxide is produced from the reaction of 48.82 g of butane and 279.5 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

$2C_{4}H_{10}+13O_{2}
ightarrow8CO_{2}+10H_{2}O$

Step2: Calculate the molar mass of butane and carbon - dioxide

The molar mass of $C_{4}H_{10}$ ($M_{C_{4}H_{10}}$) is $4\times12.01 + 10\times1.01=58.14\ g/mol$. The molar mass of $CO_{2}$ ($M_{CO_{2}}$) is $12.01+2\times16.00 = 44.01\ g/mol$.

Step3: Determine the limiting reactant

The number of moles of butane ($n_{C_{4}H_{10}}$) is $\frac{48.82\ g}{58.14\ g/mol}=0.84\ mol$. The number of moles of oxygen ($n_{O_{2}}$) is $\frac{279.5\ g}{32.00\ g/mol}=8.734\ mol$.
From the balanced equation, the mole - ratio of $C_{4}H_{10}$ to $O_{2}$ is $\frac{2}{13}$. For 0.84 mol of $C_{4}H_{10}$, the moles of $O_{2}$ required is $0.84\ mol\times\frac{13}{2}=5.46\ mol$. Since $8.734\ mol$ of $O_{2}$ is available and only $5.46\ mol$ is required, butane is the limiting reactant.

Step4: Calculate the theoretical yield of carbon - dioxide

From the balanced equation, the mole - ratio of $C_{4}H_{10}$ to $CO_{2}$ is $\frac{2}{8}=\frac{1}{4}$. So, the moles of $CO_{2}$ produced from 0.84 mol of $C_{4}H_{10}$ is $n_{CO_{2}}=0.84\ mol\times4 = 3.36\ mol$.
The theoretical mass of $CO_{2}$ ($m_{theo}$) is $m_{theo}=n_{CO_{2}}\times M_{CO_{2}}=3.36\ mol\times44.01\ g/mol = 147.87\ g$.

Step5: Calculate the percent yield

The percent yield ($\%$ yield) is given by the formula $\%$ yield=$\frac{m_{actual}}{m_{theo}}\times100\%$. Given $m_{actual} = 90.2\ g$, then $\%$ yield=$\frac{90.2\ g}{147.87\ g}\times100\% = 60.99\%\approx61.0\%$

Answer:

$61.0\%$