Give the percent yield when 14.08G of C02 are formed from the reaction of 2.000 moles...

Question

Question

Give the percent yield when 14.08G of C02 are formed from the reaction of 2.000 moles of C8H 18 with 8.000 moles of 02.

2 C8H18 + 25 02 ——> 16 C02 +18 H2O

Answer 1

Answer #1

Balanced chemical reaction is:

2 C8H18 + 25 02 ——> 16 C02 +18 H2O

moles of O2 needed to react with 2 mol of C8H18 = (25 / 2) x 2 mol = 25 mol

Here; moles of O2 reacted is = 8.0 mol (Oxygen present in limited quantity)

So; the limiting reagent is O2

moles of CO2 produced from 8.0 mol of O2 = (16 / 25) x 8 mol = 5.12 mol

Theoretical weight of CO2 = (5.12 mol) x (44.01 g/mol) = 225.33 g

practical yield of CO2 = 14.08 g

Percent yield = (14.08 / 225.33) x 100 = 6.25 %

Answer 2

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