Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
From equation;
P1V1=P2V2
V2=P1V1÷P2
since P2=380mmHg
now;1atm=760mmHg
how about 380mmHg is equal to how many atm?
380×1÷760=0.5atm
P2 now is equal to 0.5atm
back from equation;
P1V1=P2V2
V2=P1V1÷P2
V2=4.0atm×2.0L÷0.5atm
V2=16L
therefore V2=16L.
