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Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)?
When 63.1 g of HCl is allowed to react with 17.2 g of O2, 54.4 g of Cl2 is collected.
Determine the theoretical yield of Cl2 for this reaction.
- hcbiochemLv 71 month ago
63.1 g HCl / 36.46 g/mol X (2 mol Cl2 / 4 mol HCl) X 70.9 g/mol = 61.4 g Cl2
17.2 g O2 / 32.0 g/mol X (2 mol Cl2/1 mol O2) X 70.9 g/mol = 135 g Cl2
HCl is the limiting reactant, and the theoretical yield of Cl2 is 61.4 g
- Roger the MoleLv 71 month ago
4 HCl + O2 → 2 H2O + 2 Cl2
(63.1 g HCl) / (36.4611 g HCl/mol) = 1.73061 mol HCl
(17.2 g O2) / (31.99886 g O2/mol) = 0.537519 mol O2
1.73061 moles of HCl would react completely with 1.73061 x (1/4) =
0.4326525 mole of O2, but there is more O2 present than that, so O2 is in excess and HCl is the limiting reactant.
(1.73061 mol HCl) x (2 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 61.356 g =
61.4 g Cl2 in theory
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And anticipating your next question:
(54.4 g) / (61.4 g) = 0.886 = 88.6% yield Cl2