Question

gaseous butane $(ch_{3}(ch_{2})_{2}ch_{3})$ will react with gaseous oxygen $(o_{2})$ to produce gaseous carbon dioxide $(co_{2})$ and gaseous water $(h_{2}o)$. suppose 5.23 g of butane is mixed with 37. g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The balanced equation for the combustion of butane \((C_4H_{10})\) is \(2C_4H_{10}+13O_2
ightarrow8CO_2 + 10H_2O\)

Step2: Calculate the molar masses

• Molar mass of \(C_4H_{10}\): \(M_{C_4H_{10}}=(4\times12.01)+(10\times1.008)=58.12\space g/mol\)
• Molar mass of \(O_2\): \(M_{O_2}=2\times16.00 = 32.00\space g/mol\)
• Molar mass of \(CO_2\): \(M_{CO_2}=12.01+(2\times16.00)=44.01\space g/mol\)

Step3: Calculate the number of moles of reactants

• Moles of \(C_4H_{10}\): \(n_{C_4H_{10}}=\frac{m_{C_4H_{10}}}{M_{C_4H_{10}}}=\frac{5.23\space g}{58.12\space g/mol}\approx0.08999\space mol\)
• Moles of \(O_2\): \(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{37.0\space g}{32.00\space g/mol}=1.15625\space mol\)

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of \(C_4H_{10}\) to \(O_2\) is \(2:13\).
For \(0.08999\space mol\) of \(C_4H_{10}\), the moles of \(O_2\) required is \(n_{O_2}^{required}=0.08999\times\frac{13}{2}=0.5849\space mol\)
Since \(1.15625\space mol\) of \(O_2\) is available (\(1.15625>0.5849\)), \(C_4H_{10}\) is the limiting reactant.

Step5: Calculate the moles of \(CO_2\) produced

From the balanced equation, mole ratio of \(C_4H_{10}\) to \(CO_2\) is \(2:8 = 1:4\)
Moles of \(CO_2\) produced, \(n_{CO_2}=0.08999\times4 = 0.35996\space mol\)

Step6: Calculate the mass of \(CO_2\)

\(m_{CO_2}=n_{CO_2}\times M_{CO_2}=0.35996\space mol\times44.01\space g/mol\approx15.8\space g\)

Answer:

\(15.8\space g\)