
Sidney P. answered 05/31/21
Astronomy, Physics, Chemistry, and Math Tutor
a) (115g CH4) * (1 mole CH4 /16.04g CH4) * (1 mole CO2 /1 mole CH4) = 7.17 mole CO2.
(200.0g O2) * (1 mole O2 /32.00g O2) * (1 mole CO2 /1 mole O2) = 3.13 mole CO2. Oxygen is limiting.
b) (Δmole = 4.04 mole CO2) * (1 mole CH4 /1 mole CO2) * (16.04g CH4 /1 mole CH4) = 64.9g excess CH4.
c) There is not enough information to find liters of CO2 unless we are to ASSUME that we have STP. In that case, 1 mole of gas occupies 22.4 L, so 3.13 mole CO2 occupies 70.0 L of volume.