molar mass of CO2 = 12 + 2*16 = 44 grams / mol
molar mass of CO = 12 + 16 = 28 grams / mol
10 g CO = 10/28 = 0.357 mols CO in
if perfect should then get 0.357 mols CO2 out
12.2 g CO2 = 12.2/44 = 0.277 mols CO2 produced
Oh dear, got less out than perfection demands
so got .277/.357 = about 78 % of perfection
What is the percent yield of CO2 if a reaction using 10.0 g CO with excess O2 produces 12.8 g CO2?
2CO + O2 → 2CO2
2 answers
I think its 78%
2 answers