Answer:
About 58%.
Explanation:
We first need to write the balanced chemical equation:
[tex]\displaystyle \text{CH$_4$(g)} + 2\text{O$_2$(g)} \longrightarrow \text{CO$_2$(g)} + 2\text{H$_2$O(g)}[/tex]
We are given that 0.809 g of carbon dioxide was produced from the reaction of 0.80 g of methane and 2.1 g of oxygen gas. We want to calculate the percent yield of carbon dioxide.
First, determine the limiting reagent. We can convert each initial mass to the mass of carbon dioxide using stoichiometry. The molar mass of methane is 16.05 g/mol and the molar mass of carbon dioxie is 44.01 g/mol. From the equation, every one mole of carbon dioxide is produced from every one mole of methane and every two moles of oxygen gas.
0.80 g of methane will (theoretically) produce:
[tex]\displaystyle \begin{aligned} 0.80 \text{ g CH$_4$} &\cdot \frac{1 \text{ mol CH$_4$}}{16.05 \text{ g CH$_4$}} \cdot \frac{1 \text{ mol CO$_2$}}{1 \text{ mol CH$_4$}} \cdot \frac{44.01 \text{ g CO$_2$}}{1 \text{ mol CO$_2$}} \\ \\ & = 2.2 \text{ g CO$_2$}\end{aligned}[/tex]
2.1 g of oxygen gas will (theoretically) produce:
[tex]\displaystyle \begin{aligned} 2.1 \text{ g O$_2$} & \cdot \frac{ 1 \text{ mol O$_2$}}{32.00 \text{ g O$_2$}} \cdot \frac{1 \text{ mol CO$_2$}}{2 \text{ mol O$_2$}}\cdot \frac{44.01 \text{ g CO$_2$}}{1 \text{ mol CO$_2$}} \\ \\ & = 1.4 \text{ g CO$_2$} \end{aligned}[/tex]
Hence, oxygen gas is the limiting reagent. A maximum of 1.4 g of carbon dioxide can be produced.
To calculate the percent yield, we divide the actual yield by the theoretical yield. The actual yield was 0.809 g and the theoretical yield is 1.4 g. Hence:
[tex]\displaystyle \begin{aligned} \%\text{Yield} & = \frac{\text{Actual}}{\text{Theoretical}} \times 100\% \\ \\ & = \frac{(0.809\text{ g CO$_2$})}{(1.4\text{ g CO$_2$})} \times 100\% \\ \\ & = 58\%\end{aligned}[/tex]
In conclusion, the percent yield of carbon dioxide is about 58%.
