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When TiCl4 + o2 → TiO2 + Cl2rnrn109.9 grams of TiCl4 are reacted with 4.7178 moles of O2, 14.72 liters of Cl2 are produced at STP. What is the percent yield of this reaction?
Accepted Answer
The balanced chemical equation is:
TiCl4 + 2 O2 → TiO2 + 2 Cl2
Convert the mass of TiCl4 to moles:
moles TiCl4 = 109.9 g / 189.68 g/mol = 0.58 mol
Determine the limiting reactant:
moles O2 = 4.7178 mol
Ratio of moles TiCl4 to moles O2 (actual):
0.58 mol / 4.7178 mol = 0.123
Ratio of moles TiCl4 to moles O2 (stoichiometric):
1 mol / 2 mol = 0.5
Since the actual ratio is less than the stoichiometric ratio, TiCl4 is the limiting reactant.
Calculate the theoretical yield of Cl2:
moles Cl2 = 0.58 mol TiCl4 × (2 mol Cl2 / 1 mol TiCl4) = 1.16 mol
Convert the moles of Cl2 to volume at STP:
VCl2 = 1.16 mol × 22.4 L/mol = 26.0 L
Calculate the percent yield:
% yield = (actual yield / theoretical yield) × 100%
% yield = (14.72 L / 26.0 L) × 100% = 56.6%
Therefore, the percent yield of this reaction is 56.6%.
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