given the balance chemical equation, 2C2H6(g)+7O2(g)----->4CO2(g)+6H2O(g)

Question

If a reaction mixture contains 8 moles of O2(g) with 2 moles of C2H6(g),which reactant is the limiting reactqnt and what is the mass (in grams) remaining of the reactant in excess?

Answer 1

2C2H6(g) + 7O2(g) ==> 4CO2(g) + 6H2O(g)

One way to find the limiting reactant is to divide the moles of reactant by the corresponding coefficient in the balanced equation and whichever is less is limiting.

For C2H6 we have 2 moles / 2 = 1

For O2 we have 8 moles / 7 = 1.14

So in this case C2H6 would be limiting reactant and O2 would be in excess

To find the mass of O2 in excess, we can use the mol ratio of the balanced equation and dimensional analysis as follows:

moles O2 used up: 2 mol C2H6 x 7 mol O2 / 2 mol O2 = 7 moles O2 used up

moles O2 left over = 8 moles initially present - 7 moles used up = 1 mol O2 left over

mass O2 left over = 1 mol O2 x 32 g/mol = 32 g O2 left over