Question

a 10.00 l tank at 28.6 °c is filled with 8.31 g of boron trifluoride gas and 5.59 g of dinitrogen monoxide gas. you can assume both gases behave as ideal gases under these conditions. calculate the mole fraction of each gas. be sure each of your answer entries has the correct number of significant digits.

Explanation:

Step1: Calculate the molar mass of each gas

The molar mass of \(BF_3\) (\(M_{BF_3}\)): \(M_{B}=10.81\space g/mol\), \(M_{F}=19.00\space g/mol\), so \(M_{BF_3}=10.81+(3\times19.00)=67.81\space g/mol\)
The molar mass of \(N_2O\) (\(M_{N_2O}\)): \(M_{N}=14.01\space g/mol\), \(M_{O}=16.00\space g/mol\), so \(M_{N_2O}=(2\times14.01)+16.00 = 44.02\space g/mol\)

Step2: Calculate the number of moles of each gas

For \(BF_3\): \(n_{BF_3}=\frac{m_{BF_3}}{M_{BF_3}}=\frac{8.31\space g}{67.81\space g/mol}\approx0.1225\space mol\)
For \(N_2O\): \(n_{N_2O}=\frac{m_{N_2O}}{M_{N_2O}}=\frac{5.59\space g}{44.02\space g/mol}\approx0.1270\space mol\)

Step3: Calculate the total number of moles

\(n_{total}=n_{BF_3}+n_{N_2O}=0.1225 + 0.1270=0.2495\space mol\)

Step4: Calculate the mole fraction of each gas

Mole fraction of \(BF_3\) (\(x_{BF_3}\)): \(x_{BF_3}=\frac{n_{BF_3}}{n_{total}}=\frac{0.1225}{0.2495}\approx0.491\)
Mole fraction of \(N_2O\) (\(x_{N_2O}\)): \(x_{N_2O}=\frac{n_{N_2O}}{n_{total}}=\frac{0.1270}{0.2495}\approx0.509\)

Answer:

gas	mole fraction
dinitrogen monoxide	\(0.509\)