Answer to Question #122599 in General Chemistry for Marta Perehinets

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.