Question

For the reaction NH4NO3 (s) → N2O (g) + 2H2O (l), you decompose 60 g of NH4NO3 and get 11 g of N2O. What is the percent yield?
Question 8 options:

25%

33%

90%

65%

Answer 1

With your balanced equation, you know that ideally you'll get one mole of N2O for every mole of NH4NO3 that decomposes. To find out how much there should be, we need to find the moles of Ammonium Nitrate (NH4NO3) we have.
60g NH4NO3 * (1 mole NH4NO3)/(80g NH4NO3) = 0.75 moles NH4NO3
Knowing that the ratio is one to one, you should yield 0.75 moles of N2O theoretically. But, we know that we only got 11 grams of N2O, so we actually got:
11 grams N2O * (1 mole N2O)/(44 grams N2O) = 0.25 moles N2O
So we got one third of our theoretical yield, or 33%.

Answer 2

Answer: The percentage yield ids 33%.

Explanation:

$NH_4NO_3(s)\rightarrow N_2O(g)+2H_2O(l)$

Theoretical yield:

Number of moles of $NH_4NO_3$=$\frac{\text{number of moles of }NH_4NO_3}{\text{molar mass of }NH_4NO_3}=\frac{60 g}{80.04g/mol}=0.74 mol$

According to the reaction , 1 mole of $NH_4NO_3$ decomposes into 1 mole of $N_2O$ , then 0.74 moles of$NH_4NO_3$ will give 0.74 moles of $N_2O$.

Mass of $N_2O$ gas produce =

Number of moles of $N_2O$ × molar mass of $N_2O$ = 0.74 moles × 44 = 32.56 g

Theoretical yield: 32.56 g of $N_2O$

Experimental yield :

According to question ,from 60 g of $NH_4NO_3$ we get 11 g of $N_2O$

Experimental yield  = 11 g of $N_2O$

$\text{Percentage yield }=\frac{\text{experimental yield}}{\text{theoretical yield}}\times 100$

$\text{Percentage yield}=\frac{11 g}{32.56}\times 100= 33.78\%$

From the given option the closest answer to 33.78% is 33%.

Hence, the percent yield is 33%.