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How many grams of MnO, are needed to prepare 10.0 liters of Cl2, gas at 22 C and 710 torr according to the following reaction?

MnO2 + 4NaCl + 2H2SO4 2Cl2 + MnSO4 + Na2SO4 + 2H2O

User Sjrd
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1 Answer

3 votes

Answer:

15.5 g of MnO

Step-by-step explanation:

The reaction is:

MnO₂ + 2NaCl + 2H₂SO₄ → Cl₂ + MnSO₄ + Na₂SO₄ + 2H₂O

We use the Ideal Gases Law to determine the moles of produced chlorine

P . V = n . R .T

We convert T° to Abs. value → 22°C + 273 = 295K

We convert Torr to atm → 710 Torr . 1atm /760 Torr = 0.934 atm

We replace: 10 L . 0.934 atm = n . 0.082 . 295K

( 10 L . 0.934 atm ) / (0.082 . 295K) = n → 0.386 moles

As the NaCl and the acid are in excess, we assume the MnO is the limiting reactant, and ratio is 1:1. so:

0.386 moles of chlorine came from 0.386 moles of MnO

We convert moles to mass: 0.386 . 40.3g/mol = 15.5 g

User Santiago Varela
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