Question

a mixture of 3.86 g of ccl4 (carbon tetrachloride) and 1.92 g of c2h4 (ethylene) at 450 c exert how many atmospheres of pressure inside a 30- ml metal bomb? how much pressure is contributed by ethylene?

Answer 1

First calculate the total number of moles of mixture as follows:

Number of mole ccl4 = 3.86 g of ccl4 / 153.82 g/ mol = 0.025 mol

Number of mole c2h4 (ethylene) = 1.92 g of C2H4 / 28.05 g/ mol = 0.068 mol

Total number of moles = 0.025 mol +0.068 mol = 0.093 mol

Now use the gas equation to calculate the pressure inside the bomb as follows:

PV=nRT

V= 30- ml or 0.03 L

R= 0.0821 L atm / mole –K

T= 450 c or 723 K

Then P , pressure calculated as follows:

P = nRT /V

P = 0.093 mol *0.0821 L atm / mole –K*723 K /0.03 L

P= 184 atm