The reaction of 5.78g of Carbon with excess O2 yields 13.45g of CO2. What is the percent yield?
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C+ O2 ----------------- CO2
MOLES of carbon that reacted =
moles = mass in grams/ molar mass
5.78/ 12 = 0.4817 moles
from formula moles of carbon reacted = moles of CO2 formed
moles of CO2 expected to be formed = 0.4817
theoretical yield = moles * molar mass
= 0.4817 * 44
% yield = actual yield/ theoretical yield * 100%
actual yield = 13.45g
% yield = 13.45/ 21.193 * 100%
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