Question

$0.4\, g$ of dihydrogen is made to react with $7.1 \,g$ of dichlorine to form hydrogen chloride. The volume of hydrogen formed at $273\,K$ and $1$ bar pressure is

• A. 9.08L
• B. 4.54L
• C. 90.8L
• D. 45.4L

Hint:

The reaction between dihydrogen and dichlorine is: \[ H_2 + Cl_2 \rightarrow 2HCl \] We use the ideal gas law to calculate the volume of hydrogen chloride formed. First, we need to find the limiting reagent.
Molar mass of dihydrogen (\(H_2\)) = 2 g/mol, Molar mass of dichlorine (\(Cl_2\)) = 71 g/mol.
Number of moles of \(H_2\) = \(\frac{0.4}{2} = 0.2 \, mol\), Number of moles of \(Cl_2\) = \(\frac{7.1}{71} = 0.1 \, mol\). From the stoichiometry, 1 mole of \(H_2\) reacts with 1 mole of \(Cl_2\), so \(Cl_2\) is the limiting reagent. Therefore, 0.1 moles of \(Cl_2\) will produce 0.2 moles of \(HCl\). Now, using the ideal gas equation \(PV = nRT\):
\[ V = \frac{nRT}{P} \] Where: \(n = 0.2 \, mol\), \(R = 0.0831 \, \text{L.bar/(mol.K)}\), \(T = 273 \, K\), \(P = 1 \, bar\). Substituting the values: \[ V = \frac{0.2 \times 0.0831 \times 273}{1} = 4.54 \, L \]

So, the correct answer is (B): \(4.54\ L\)

Answer 1

The Correct Option is B

The corresponding balanced reaction is:

$H _{2}+ Cl _{2} \rightarrow 2 HCl$

Again, $0.4 \,g H _{2}=0.4 / 2 \,mol =0.2 \,mol$
$7.1 \, g Cl _{2}=7.1 / 71 \,mol =0.1 \,mol$

So, $Cl _{2}$ is the limiting reagent here.

Hence $0.2$ moles of $HCl$ will be formed.

Again, we need to find out the volume of $HCl$ formed in the STP condition.

1 mol of gas in STP occupies $22.7 \, L$

Thus, $0.2$ mol of $HCl$ will occupy $=4.54 \, L$