Question

`"450 g"` of hydrogen gas reacts with an excess of nitrogen gas to produce `"15.75 g"` of ammonia. What is the percent yield?

Answer 1

The percent yield of ammonia is `"11%"`.

Explanation:

In order to determine percent yield, the theoretical yield must first be calculated. This is done by converting the mass of the limiting reactant to moles, then multiplying times the mole ratio from the balanced equation that has ammonia on top and hydrogen on the bottom. Once the theoretical yield is calculated, the percent yield is determine by the following equation:

`"percent yield"` = `"actual yield"/"theoretical yield"` x `"100"`

Start with a balanced equation:

`"N"_2("g")` + `"3H"_2("g")` `rarr` `"2NH"_3("g")`

Since nitrogen is in excess, the limiting reactant is hydrogen. The molar mass of `"H"_2("g")` is `"2.016g/mol"`.

Convert the mass of `"H"_2` to moles.

`450color(red)cancel(color(black)("g H"_2))xx(1"mol H"_2)/(2.016color(red)cancel(color(black)("g H"_2)))= "223.2 mol H"_2`

Determine the theoretical yield of ammonia by multiplying mol `"H"_2"` by the mole ratio between ammonia and hydrogen from the balanced equation, with ammonia in the numerator.

`223.2 color(red)cancel(color(black)("mol H"_2))xx(2 "mol NH"_3)/(3 color(red)cancel(color(black)("mol H"_2)))="148.8g NH"_3`

The theoretical yield of ammonia in this reaction is `"148.8g"`

The actual yield is `"15.75g NH"_3`.

The percent yield is `"15.75g NH"_3/"148.8g NH"_3` x `"100"` = `"11"` (rounded to two significant figures due to `"450g"` having only two significant figures.