Question

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

Answer 1

Hint: At given Standard Temperature and Pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L. According to the Ideal Gas Law, along with a balanced chemical equation this law can be used to solve for the amount, either in volume or mass, of gas consumed or produced in a chemical reaction.

Complete step by step answer:
Given in question:
Mass of elemental boron = 21.6 g
Temperature of the gas = 273 K
Pressure of the gas = 1 atm
The balanced chemical equation for the reduction of boron dichloride is given as follows:
\[BC{{l}_{3}}+\dfrac{3}{2}{{H}_{2}}\to B+3HCl\]
From the equation, we can clearly see that 1 mole of boron formation requires 1.5 moles of hydrogen gas to react and form 1 mole of elemental boron and 3 moles of hydrogen chloride.
Now let us calculate the moles of boron from the given mass of boron.
Number of moles of boron can be given by dividing the given mass of boron by the molar mass.​
Number of moles of boron= \[\dfrac{21.6}{10.8}\]mol
= 2.00 mol


So, to obtain 2.00 mol of elemental boron, 3.00 mol of hydrogen gas is required, by unitary method.
The volume occupied by one mole of hydrogen gas at 273 K and 1 atm is the standard volume which is 22.4L.
Therefore, volume of 3 moles hydrogen gas at 273 K temperature and 1 atm pressure = 3.00 × 22.4 = 67.2L

Note:
Reduction is defined as a chemical reaction that involves the gaining of electrons by one of the atoms involved in the reaction between two chemical compounds. The term refers to the element that accepts electrons and experiences a decrease in the oxidation state. As the oxidation state of the element that gains electrons is lowered it is called reduction.