Question

A reaction to convert ammonia into nitric acid involves the following chemical equation: 

            4 NH3 (g) + 5 O2 (g) 4 NO(g) + 6 H2O(g)

 Assume that 1.50 g of NH3 reacts with 2.75 g O2. 

i. Determine the limiting reactant.

ii. Calculate how many grams of NO and H2O will form in the reaction above.

iii. Calculate how many grams of the excess reactant will remain after the reaction is complete. 

iv. If the yield obtained for NO is 1.80 grams and yield obtained for H2O is 1.50 grams, calculate the percent yield of NO and H2O in this reaction.

        (Atomic mass of N = 14 ; O = 16 ; H = 1)

Answer 1

Moles of Ammonia = $\frac{1.50}{17}=0.088 mol$

Moles of Oxygen$=\frac{2.75}{32}=0.086 mol$

$Difference=0.088−0.086=2.0×10^{−3}mol$

Mass of excess reactant that remain = $[2.0×10^{-3}]×17=0.034grams$

Moles of NO $=\frac{1.80}{30}=0.06 moles$

Moles of H2O=$=\frac{1.50}{33}=0.045 moles$

Limiting reactants is NH₃