4.35 L of N2, at 25 ∘C and 1.16 atm, is mixed with 2.20 L of O2, at 25 ∘C and 0.277 atm, and the mixture allowed to react. N2 (g) + O2 (g) →2NO (g) How much NO, in grams, is produced?

Mass = 1.5 g

Explanation:

Given dа ta:

Volume of nitrogen = 4.35 L

Temperature = 25°C

Pressure = 1.16 atm

Volume of oxygen = 2.20 L

Temperature = 25°C

Pressure = 0.277 atm

grams of NO = ?

Solution:

Moles of Nitrogen:

PV = nRT

n = PV/RT

n = 1.16 atm*4.35 L / 0.0821 atm. L. K⁻¹. mol⁻¹ * 298 K

n = 5.046 atm. L / 24.47 atm. L. mol⁻¹

n = 0.206 mol

Moles of oxygen:

PV = nRT

n = PV/RT

n = 0.277 atm*2.20 L / 0.0821 atm. L. K⁻¹. mol⁻¹ * 298 K

n = 0.609 atm. L / 24.47 atm. L. mol⁻¹

n = 0.025 mol

Balanced chemical equation:

N₂ + O₂ → 2NO

Now we will compare the moles of NO with moles of oxygen and nitrogen.

N₂ : NO

1 : 2

0.206 : 2*0.206 = 0.412 mol

O₂ : NO

1 : 2

0.025 : 2*0.025 = 0.05 mol

The number of moles produced by oxygen are less so it will limiting reactant.

Mass of NO:

Mass = number of moles * molar mass

Mass = 0.05 mol * 30 g/mol

Mass = 1.5 g