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 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