Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H 2O(l)+Cl2(g)A sample of 36.9 g MnO2 is added to a solution containing 48.9 g HCl. What is the limiting reactant?MnO2HClWhat is the theoretical yield of Cl2?theoretical yield:g Cl2If the yield of the reaction is 74.7%, what is the actual yield of chlorine?actual yield:g Cl2

HCl is limiting reactant

Theoretical yield: 23.8g Cl₂

Actual yield: 17.6g C₂

Explanation:

Based on the reaction:

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.

To find the limiting reactant you must know the moles of each reactant and knowing that 4 moles of HCl reacts per mole of MnO₂ you can sikve this problem, thus:

Moles HCl (Molar mass: 36.46g/mol): 48.9g ₓ (1mol / 36.46g/mol) =

1.341 moles HCl

Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =

0.424 moles MnO₂

For a complete reaction of 0.424 moles of MnO₂ you require:

0.424moles MnO₂ ₓ (4 moles HCl / 1 mole MnO₂) = 1.696 moles of HCl.

As you have just 1.341 moles of HCl. HCl is limiting reactant.

Theoretical yield means, in the reaction, that 4 moles of HCl will produce 1 mole of Cl₂. As moles of HCl are 1.341:

1.341 moles HCl ₓ (1 mole Cl₂ / 4 moles HCl) = 0.33525 moles Cl₂

In grams (Molar mass Cl₂: 70.9g/mol):

Theoretical yield: 0.33525 moles Cl₂ ₓ (70.9g / mol) = 23.8g Cl₂

As yield of reaction is 74.7%, the real mass of Cl₂ you obtain (Actual yield) is:

23.8g Cl₂ ₓ 74% = 17.6g C₂

it's filtration, but this is an assumption so my apologies if i got this

the answer would be b i believe