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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 limiting reactant for this reaction.
Express your answer as a chemical formula.
- Roger the MoleLv 72 months 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.