If you start with 85g of Nitrogen and isolate 86g of ammonia, what is your percent yield?

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g)

moles N₂ used = 85 g N₂ x 1 mol/28 g = 3.04 moles N₂

moles NH₃ possible at 100% yield = 3.04 moles N₂ x 2 moles NH₃/mol N₂ = 6.08 moles NH₃

mass of NH₃ = 6.08 moles NH₃ x 17 g/mole = 103 g

percent yield = actual yield/theoretical yield (x100%) = 86 g/103 g (x100%) = 83.5% = 84% yield

Answer 2

Start by writing the chemical reaction for nitrogen to ammonia

N + 3H —> NH₃

_{From the equation you can see that for every mole of nitrogen you get one mole of ammonia}

Next, calculate the moles of nitrogen in 85g

85g • (1mole/14g) = 6.07 moles of Nitrogen

Next calculate the expected the yield

6.07 moles of nitrogen •(1 mole of ammonia / 1 mole nitrogen) = 6.07 moles of ammonia

Convert the moles of ammonia to grams of ammonia

6.07 moles of ammonia • (17.03 g / 1 mole of ammonia) = 103.396 g of ammonia

Last, calculate the yield percent by dividing the mass you got over the mass you could have possibly gotten and multiply by 100 to get the percent

(86 g / 103.396 g) • 100 = 83.18%