Question

A reaction to convert ammonia into nitric acid involves following chemical

equation:

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

Assume that 1.50 g of NH3 reacts with 2.75 g O2, then:

(a) Determine the limiting reactant.

(b) Calculate how many grams of NO and H2O will form in the reaction above.

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