Question

if 87 g of nitrogen monoxide, no, is produced from the reaction of 150 g o₂ with excess nh₃, what is the percentage yield? (molar masses: o₂ = 32.0 g/mol, no = 30.0 g/mol) 4nh₃(g) + 5o₂(g) → 4no(g) + 6h₂o(g) 49% 33% 77% 58% 62%

Explanation:

Step1: Calculate moles of \(O_2\)

Moles of \(O_2=\frac{mass}{molar\ mass}=\frac{150\ g}{32.0\ g/mol}\approx4.6875\ mol\)

Step2: Find moles of \(NO\) using stoichiometry

From \(4NH_3 + 5O_2
ightarrow4NO + 6H_2O\), mole ratio \(O_2:NO = 5:4\)
Moles of \(NO=\frac{4}{5}\times4.6875\ mol = 3.75\ mol\)

Step3: Calculate theoretical mass of \(NO\)

Theoretical mass of \(NO=moles\times molar\ mass = 3.75\ mol\times30.0\ g/mol = 112.5\ g\)

Step4: Calculate percentage yield

Percentage yield\(=\frac{actual\ yield}{theoretical\ yield}\times100=\frac{87\ g}{112.5\ g}\times100\approx77\%\)

Answer:

77 %