= 24 mol CO2
= 10 mol CO2
When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction,
2 CH3OH(g) + 3 O2 (g) → 2 CO2 (g) + 4 H2O(g)
what is the excess reactant and how many moles of it remain at the end of the reaction?
a) 9 mol CH3OH (g)
b) 10 mol CO2 (g)
c) 10 mol CH3OH (g)
d) 14 mol CH3OH (g)
e) 1 mol O2 (g)
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