Given
Find
Solution
\(
\begin{array}{l}
C_{4}H_{10}\ (g) + 6,5\ O_{2}\ (g) \to 4\ CO_{2}\ (g) + 5\ H_{2}O\ (g) \\\\
M(C_{4}H_{10}) \approx 4 \cdot 12\frac{g}{mol} + 10 \cdot 1\frac{g}{mol} = 58\frac{g}{mol} \\
M(CO_{2}) \approx 12\frac{g}{mol} + 2 \cdot 16\frac{g}{mol} = 44\frac{g}{mol} \\
M(H_{2}O) \approx 2 \cdot 1\frac{g}{mol} + 16\frac{g}{mol} = 18\frac{g}{mol} \\\\
\frac{n(C_{4}H_{10})}{n(CO_{2})} = \frac{1}{4},\ \frac{n(C_{4}H_{10})}{n(H_{2}O)} = \frac{1}{5} \\\\
n(C_{4}H_{10}) = \frac{m(C_{4}H_{10})}{M(C_{4}H_{10})} \approx \frac{10g}{58\frac{g}{mol}} \approx 0,1724mol \\
n(CO_{2}) = 4 \cdot n(C_{4}H_{10}) \approx 0,6896mol\\
n(H_{2}O) = 5 \cdot n(C_{4}H_{10}) \approx 0,862mol \\\\
m(CO_{2}) = M(CO_{2}) \cdot n(CO_{2}) \approx 30,3424g \\
m(H_{2}O) = M(H_{2}O) \cdot n(H_{2}O) \approx 15,516g \\
\end{array}
\)
Answer