Question

Question #7d102

Answer 1

Your theoretical yield will be `"32.9 g"` and the reaction's percent yield will be `"81.2%"`.

Once again, start with the balanced chemical equation

`N_(2(g)) + 3H_(2(g)) -> 2NH_(3(g))`

Notice the `"3:2"` mole ratio that exists between hydrogen gas and ammonia; this means that for every 3 moles of hydrogen gas that react, 2 moles of ammonia will be produced.

In other words, regardless of how many hydrogen moles react, you'll always have 2/3 less moles of ammonia produced.

Determine the number of moles of hydrogen gas by using its molar mass

`"5.84 g H"_2 * "1 mole H"_2/"2.016 g" = "2.897 moles H"_2`

For a 100% yield, all the moles of hydrogen gas must react to produce ammonia. This means that the moles of ammonia produced will be

`"2.897 moles H"_2 * "2 moles NH"_3/"3 moles H"_2 = "1.931 moles NH"_3`

Use ammonia's molar mass to see how many grams would be produced in a 100%-yield reaction `->` this is your theoretical yield.

`"1.931 moles NH"_3 * "17.03 g"/"1 mole NH"_3 = "32.9 g NH"_3`

However, your reaction produces less ammonia (26.7 g) than what was calculated for a 100% yield, which means that your reaction's percent yield will be smaller than 100%.

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

`"% yield" = "26.7 g"/"32.9 g" * 100 = "81.2%"`