Question

For the reaction shown, calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g)

a. 6.4 mol N2O5

b. 10.0 g N2O5

c. 1.55 kg N2O5

Answer 1

2 N₂O₅ (g)    →   4 NO₂ (g) + O₂ (g)

Looking at the chemical equation, 2 moles of N₂O₅ = 4 moles of NO₂

a. ((6.4 mol N₂O₅/1) X 4 mol NO₂)/2 mol N₂O₅

6.4 X 4 = 25.6

25.6/2 =12.8

12.8 mol NO₂ are formed

b. 10g N₂O₅ / 108.01g/mol = 0.0926 mol N₂O₅

((0.0926 mol N₂O₅/1) X 4 mol NO₂)/2 mol N₂O₅

0.0926 X 4 = 0.3704

0.3704/2 =0.1852

0.1852 mol NO₂ are formed

a. 1.55kg N₂O₅ = 1550g N₂O₅

1550 g N₂O₅ / 108.01 g/mol N₂O₅ = 14.35mol N₂O₅

((14.35 mol N₂O₅/1) X 4 mol NO₂)/2 mol N₂O₅

14.35 X 4 = 57.4

57.4/2 =28.7

28.7 mol NO₂ are formed