Question

What volume of hydrogen gas, at 273 k and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?

Answer 1

Answer: Volume of hydrogen gas consumed will be 67.20 L.

Explanation: Moles of Boron is calculated by:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of boron = 21.6g

Molar mass of boron = 10.8g/mol

Putting values in above equation, we get

$Moles=\frac{21.6g}{10.8g/mol}$

Moles of boron = 2 moles

Reduction of boron trichloride by hydrogen is given by the equation:

$2BCl_3+3H_2\rightarrow 2B+6HCl$

By Stoichiometry, we get:

2 moles of boron are produced by 3 moles of hydrogen gas,so

2 moles of boron will be produced by = $\frac{3}{2}\times 2=3moles$ of hydrogen gas.

Using Ideal gas equation, which is:

$PV=nRT$

Given:

P = 1 atm

T = 273K

n = 3 moles

$R=0.082057\text{ L atm }mol^{-1}K^{-1}$  (Gas Constant)

Putting all the values in above equation, we calculate the volume consumed:

$1atm\times V=(3mol)(0.082057\text{ L atm }mol^{-1}K^{-1})(303K)$

V = 67.20 L

Answer 2

Answer: Volume of hydrogen gas consumed will be 67.20 L.

Explanation: Moles of Boron is calculated by:

Given mass of boron = 21.6g

Molar mass of boron = 10.8g/mol

Putting values in above equation, we get

Moles of boron = 2 moles

Reduction of boron trichloride by hydrogen is given by the equation:

By Stoichiometry, we get:

2 moles of boron are produced by 3 moles of hydrogen gas,so

2 moles of boron will be produced by =  of hydrogen gas.

Using Ideal gas equation, which is:

Given:

P = 1 atm

T = 273K

n = 3 moles

 (Gas Constant)

Putting all the values in above equation, we calculate the volume consumed:

V = 67.20 L