Answer:
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)
<em>4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.</em>
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) =
<em>1.341 moles HCl</em>
Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =
<em>0.424 moles MnO₂</em>
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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
