Question #7d102

1 Answer
Mar 14, 2015

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%"#