Question

If 0.632g of NH3(g) are allowed to react with 0.438g of O2(g), calculate the mass of
NO(g) that would be produced in the following reaction: (+ identify limiting reagent)
4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)

Answer 1

4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)

0.632/17 x 4 = 0,00929411765 moles of NH3.

0.438/32 x 5 = 0,0027375 moles of O2.

0.0027375 x 4 x 30 = 0,3285 g of NO.