Calculate the volume of Cl2 gas (in mL) liberated at 1 atm & 273 K when 1.74 g MnO2 reacts with 2.19 g HCl according to the following reaction with % yield 40. MnO2 + HCl → MnCl2 + Cl2 + H2O

Engineering

Chemistry

Chemical Equation and Stoichiometric Calculation

Question

Calculate the volume of Cl₂ gas (in mL) liberated at 1 atm & 273 K when 1.74 g MnO₂ reacts with 2.19 g HCl according to the following reaction with % yield 40.

MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O

112 mL

336 mL

44.8 mL

134.4 mL

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MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

1.74 g 2.19 g

moles 0.02 0.06
to find LR $\frac{0 .02}{1}$ $\frac{0 .06}{4}$

HCl is limiting reagent

4 mole of HCl gives 1 mole of Cl₂

1 → 1/4

0.06 → 1/4 × 0.06 mole of Cl₂ = 0.015

$% yield = \frac{Actual product formed}{theoritical product formed} \times 100$

$40 = \frac{x}{0 .015} \times 100$
$x = 40 \times \frac{0 .015}{100} = 0 .006 moles {Cl}_{2} actual$
formed at 1 atm, 273 K = 0.006 × 22.4 (lit.) = 134.4 ml