MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + H2O How many grams of HCl are required to produce 11.2 liters of Cl2? The reaction occurs in the presence of excess HCl and at STP. A) 20.0 g B) 36.5 g C) 48.5 g D) 73.0 g

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28-05-2020
Chemistry

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MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + H2O How many grams of HCl are required to produce 11.2 liters of Cl2? The reaction occurs in the presence of excess HCl and at STP. A) 20.0 g B) 36.5 g C) 48.5 g D) 73.0 g

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mickymike92 mickymike92 · Answer 1 · 2020-05-28T04:44:33+02:00

mickymike92 mickymike92

28-05-2020

Answer:

73.0g of HCl

Explanation:

Check the attachment below for explanation.

Answer Link

Abdulazeez10 Abdulazeez10 · Answer 2 · 2022-01-27T11:57:40+01:00

The mass of HCl required to produce 11.2 liters of Cl₂ at STP is 73.0 g. The correct option is D) 73.0 g

Stoichiometry

From the question,

We are to determine the mass of HCl required to produce 11.2 liters of Cl₂

From the given balanced chemical equation,

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + H₂O

This means,

1 mole of MnO₂ reacts with 4 moles of HCl to produce 1 mole of Cl₂

Now, we will determine the number of moles of Cl₂ that is required

Volume of Cl₂ required = 11.2 L

NOTE: At STP, 1 mole of an ideal occupies 22.4 L

If 1 mole of a gas occupies 22.4 L

Then,

0.5 mole of Cl₂ will occupy 11.2 L

Therefore, the number of moles of Cl₂ required to be produced is 0.5 moles

Now,

From the balanced chemical equation,

If 4 moles of HCl are required to produce 1 mole of Cl₂

Then,

2 moles of HCl will be required to produce 0.5 mole of Cl₂

For the mass of HCl required,

Using the formula,

Mass = Number of moles × Molar mass

Number of moles of HCl = 2 moles

Molar mass of HCl = 36.46 g/mol

Thus,

Mass of HCl required = 2 × 36.46

Mass of HCl required = 72.92 g

Mass of HCl required ≅ 73 g

Hence, the mass of HCl required to produce 11.2 liters of Cl₂ at STP is 73.0 g. The correct option is D) 73.0 g

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