# Question #39a18

Mar 14, 2015

The theoretical yield will be $\text{7.48 g}$ of carbon dioxide and the reaction's percent yield will be $\text{93.9%}$.

$2 C {O}_{\left(g\right)} + {O}_{2 \left(g\right)} \to 2 C {O}_{2 \left(g\right)}$

Notice that you have a $\text{1:1}$ (2:2) mole ratio between carbon monoxide and carbon dioxide - more precise, for every mole of the former that reacts, 1 mole of the latter will be produced.

This is what your theoretical yield will be $\to$ the number of moles of $C {O}_{2}$ produced must be equal to the number of moles of $C O$ that reacted, assuming in this case that all available moles of $C O$ will react.

Determine the number of moles of $C O$ by using its molar mass

$\text{4.76 g CO" * "1 mole CO"/"28.01 g" = "0.1699 moles CO}$

According to the mole ratio that exists between the two compounds, the number of moles of $C {O}_{2}$ produced for a 100% yield will be

${\text{0.1699 moles CO" * "1 mole CO"_2/"1 mole CO" = "0.1699 moles CO}}_{2}$

Use $C {O}_{2}$'s molar mass to see how many grams should have been produced

$\text{0.1699 moles CO"_2 * "44.0 g"/"1 mole CO"_2 = "7.48 g}$

This is your theoretical yield. But since you've actually produced 7.02 g of $C {O}_{2}$, your percent yield will be less than 100%.

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

$\text{% yield" = "7.02 g"/"7.48 g" * 100 = "93.9%}$