What pressure is exerted by a mixture of 2.80 g of H2 and 6.100 g of N2 at 273°C in a 10.0 L container?

Pressure exerted is 7.24 atm

Explanation:

Let's apply the Ideal Gases Law:

P . V = n . R . T

First of all, we must convert the mass to moles, to know the total moles of the mixture:

2.8 g . 1 mol / 2g = 1.4 moles of H

6.1 g . 1mol / 28 g = 0.218 moles of N

Total moles = 1.4 + 0.218 → 1.618 moles

P . 10 L = 1.618 moles . 0.082 L.atm/mol.K . 546K

P = (1.618 moles . 0.082 L.atm/mol.K . 546K) / 10 L

P = 7.24 atm

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