Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2?

Question

with 29.0 L of CO2 (at STP).

4 KO2(s) + 2 CO2(g) ---> 2 K2CO39s) + 3 O2(g)

I need some help breaking this down, step by step.

Answer 1

27.9 g KO2 @ 71.10 g/mol = 0.3924 molesKO2

29.0 L stp @ 1 mole / 22.4 L stp = 1.2946 moles CO2

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at this ratio: 4 KO2(s) + 2 CO2(g) ---> 2 K2CO3(s) + 3 O2(g)

they have added quite an excess of CO2, so the KO2 is the limiting reagent

? mol K2CO3 = 0.3924 mol KO2 @ 2 mol K2CO3 /4 mol K2O = 0.1962 mol K2CO3 is max yield

0.1962 mol K2CO3  @ 138.2 g/mol = 27.11 g K2CO3 expected

21.8 g / 27.11 g K2CO3  = 80.3988

kyour answer: 80.4% yield