Answer to Question #165635 in General Chemistry for Maddy Brown

Question

Consider the balanced chemical equation shown below.

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

In a certain experiment, 4.699 g of NH3(g) reacts with 1.850 g of O2(g).

A) Which is the limiting reactant?

B) How many grams of NO(g) form?

C) How many grams of H2O(g) form?

D) How many grams of the excess reactant remains after the limiting reactant is completely consumed?

Answer 1

$4NH_{3}(g) + 5O_{2}(g)\rightarrow4NO(g)+6 H_{2}O(g)$

(a) According to stoichiometry,

5 moles of O₂ reacts with 4 moles of NH₃

160 g of O₂ reacts with 68 g of NH₃

1.850 g of O₂ reacts with = $\dfrac{68}{160}\times1.850$ g of NH₃

= $0.78625$ g of NH₃

Since mass of O₂ is fully consumed in the reaction and 3.91275 g of NH₃ is in excess

$\therefore$ O₂ is the limiting reactant.

(b) 5 moles of O₂ reacts to form 4 moles of NO

160 g of O₂ react to form 132 g of NO

$\therefore$ 1.850 g of O₂ react to form = $\dfrac{132}{160}\times 1.850$ g of NO

= $1.52625$ g of NO

(c) Similarly mass of H₂O can be found using stoichiometry,

Mass of H₂O formed = $\dfrac{108}{160}\times1.850$

= $1.24875$ g

(d) Mass of NH₃(excess reactant) left = Total mass - consumed mass

= $4.699-0.78625$

= $3.91275$ g