Question

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?

Answer 1

Mass = 1.5 g

Step-by-step explanation:

Given data:

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