Question

Question #39a18

Answer 1

The theoretical yield will be `"7.48 g"` of carbon dioxide and the reaction's percent yield will be `"93.9%"`.

Start with the balanced chemical equation

`2CO_((g)) + O_(2(g)) -> 2CO_(2(g))`

Notice that you have a `"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 `->` the number of moles of `CO_2` produced must be equal to the number of moles of `CO` that reacted, assuming in this case that all available moles of `CO` will react.

Determine the number of moles of `CO` by using its molar mass

`"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 `CO_2` produced for a 100% yield will be

`"0.1699 moles CO" * "1 mole CO"_2/"1 mole CO" = "0.1699 moles CO"_2`

Use `CO_2`'s molar mass to see how many grams should have been produced

`"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 `CO_2`, your percent yield will be less than 100%.

`"% yield" = "actual yield"/"theoretical yield" * 100`

`"% yield" = "7.02 g"/"7.48 g" * 100 = "93.9%"`