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A sample of 39.1 g MnO₂ is added to a solution containing 45.3 g HCl.

1. What is the limiting reactant?
2. What is the theoretical yield of Cl₂?
3. If the yield of the reaction is 79.1%, what is the actual yield of chlorine?

Answer :

Final answer:

The limiting reactant is MnO2. The theoretical yield of Cl2 is 0.2245 mol. The actual yield of chlorine is 0.1776 mol.

Explanation:

To determine the limiting reactant, we need to compare the number of moles of each reactant with their stoichiometric coefficients. The molar mass of MnO2 is 86.94 g/mol, so the number of moles is 39.1 g / 86.94 g/mol = 0.449 mol. The molar mass of HCl is 36.46 g/mol, so the number of moles is 45.3 g / 36.46 g/mol = 1.242 mol. Since MnO2 has a lower number of moles, it is the limiting reactant.



The balanced equation for the reaction is:



2 MnO2 + 4 HCl → 2 MnCl2 + Cl2 + 2 H2O



From the balanced equation, we can see that 2 moles of MnO2 produce 1 mole of Cl2. Therefore, the theoretical yield of Cl2 is 0.449 mol x 1 mol/2 mol = 0.2245 mol.



To find the actual yield of chlorine, we need to multiply the theoretical yield by the percent yield. The actual yield = (percent yield / 100) x theoretical yield = (79.1 / 100) x 0.2245 mol = 0.1776 mol.

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