Question

What is the total pressure exerted by a mixture of 48.0 grams of CH4 and 56.0 grams of
hydrogen gas when confined in a volume of 15.0 liters at 13°C.

Answer 1

The pressure of the gas is obtained as 48 atm.

What is the total pressure?

Now we know that;

Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles

Number of moles of H2 =  56.0 grams/2 g/mol = 28 moles

Total number of moles present = 3 moles + 28 moles = 31 moles

Using;

PV =nRT

P = total pressure

V = total volume

n = total number of moles

R = gas constant

T = temperature

P = nRT/V

P = 31 * 0.082 * 286/15

P = 48 atm

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