Consider a situation in which 211 g of P4 are exposed to 240 g of O2....
Consider a situation in which 211 g of P4 are exposed to 240 g of O2.
P4+5O2→2P2O5
Maximum number of moles of P2O5 that can theoretically be made from 211 g of P4 and excess oxygen = 3.4 mol
Maximum number of moles of P2O5 that can theoretically be made from 240 g of O2 and excess phosphorus = 3 mol
Question: What is the percent yield if the actual yield from this reaction is 341 g ?
Express your answer to three significant figures and include the appropriate units.
Homework Answers
According to your calculations oxygen is limiting reagents so product will be decided by the amount of O2
Moles of P4O10 = 3
Mass of P4O10 = mole * molar mass
= 3* 284
= 852 grams
% yield = ( actual mass / calculated mass ) *100
= 341 *100 / 852
= 40.0 %
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