Question

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide. 2 NO(g) + O2(g) → 2 NO2(g)(a) You wish to react NO and O2 in the correct stoichiometric ratio. The sample of NO has a volume of 150 mL. What volume of O2 is required (at the same pressure and temperature)? (b) What volume of NO2 (at the same pressure and temperature) is formed in this reaction?

Answer 1

Step-by-step explanation:

a) To solve this question we need to know that the molar volume occupied by any gas at STP is equal to 22.4 L/mol.

So in this case we have the following reaction:

2 NO(g) + O2(g) → 2 NO2(g)

The stoichiometric ratio between NO and O2 is 2:1.

If we have 150 mL of NO, we can find the value into moles using the molar volume:

22.4 L ---- 1 mol

0.150 L ---- x mol

x = 0.150/22.4

x = 6.7 x 10^-3 moles NO

using the stoichiometric ratio between NO and O2 we can find the quantity on moles of O2:

2 moles of NO --- 1 mole of O2

6.7 x 10^-3 moles NO ---- x mole of O2

x = 3.3 x 10^-3 moles of O2

Now let's transform to liters:

22.4 L ---- 1 mol

x L ---- 3.3 x 10^-3 mol

x = 0.075 L = 75 mL

b) It is formed 150 mL of NO2, because the stoichiometric ratio between NO and NO2 is 2:2, so it will form the same amount of NO used.

Answer: a) 75 mL of O2

b) 150 mL of NO2