What is the total pressure exerted by a gaseous mixture that consists of 8.00 g of methane and 12.00 g of ethane, C2H6, in a 3.50 L container maintained at 35.20 °C?

A) 6.49 atm B) 0.400 atm C) 3.13 atm D) 16.5 atm E) 0.741 atm

**Solution:**

For this question we need to use the **Ideal Gas Law: PV= nRT **where P is pressure in atm, V is volume in L, n is the number of moles, R is the ideal gas constant and T is temperature in Kelvin.

P = ?

V = 3.50L

n = total moles

R = 0.08206 L*atm/mol*K

T = 35.20 °C+273 = 308.2K

Let's find the total number of moles of both gases.

methane = 8.00g/16.04g/mol = 0.49875mol

ethane = 12.00g/ 30.07 g/mol = 0.3991 mol

Total moles = 0.49875mol+0.3991 mol = 0.8979mol

We can now plug in all the values into the equation,

P(3.50L)=0.8979mol* 0.08206 L*atm/mol*K*308.2K

P = 6.49 atm

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