Question #18238
calculate the mass of butane needed to produce 53.9 grams of carbon dioxide
Expert's answer
2C4H10+ 13O2 = 8CO2 + 10H2O;
n(CO2) = 53.9g/44 g/mol
n(CO2) = 1.225 mol;
n(C4H10) = 0.30625 mol;
m(C4H10) = 0.30625 mol * 58 g/mol = 17.7625 g
n(CO2) = 53.9g/44 g/mol
n(CO2) = 1.225 mol;
n(C4H10) = 0.30625 mol;
m(C4H10) = 0.30625 mol * 58 g/mol = 17.7625 g
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
#340153 on Dec 2023
#340153 on Dec 2023