Assuming that alcohol is ethanol, the reaction is following:
C2H5OH + 3O2 --> 2CO2 + 3H2O
a) 3 moles of O2 are needed for combustion of 1 mole of alcohol. More details cannot be determined since the amount of alcohol is not specified.
b) 1 mole
c) m(CO2)=1 mol(C2H5OH)×1 mol(C2H5OH)2 mol(CO2)×1 mol(CO2)44.01 g(C2)=88 g
Answer: 88 g
d) m(CO2)=1 g(C2H5OH)×46.07 g(C2H5OH)1 mol(C2H5OH)×1 mol(C2H5OH)2 mol(CO2)×1 mol(CO2)44.01 g(C2)=1.9 g
Answer: 1.9 g
3.
The reaction is below:
C6H12O6 --> 2C2H5OH + 2CO2
m(C2H5OH)=2000 g(C6H12O6)×180.2 g(C6H12O6)1 mol(C6H12O6)×1 mol(C6H12O6)2 mol(C2H5OH)×1 mol(C2H5OH)46.07 g(C2H5OH)=1023 g
Answer: 1023 g
Comments