C3H8(g)+5O2(g)-> 3CO2(s)+4H2O(g)

Question

Given the equation above how many moles of H20 gas are produced when 7.9 moles propane (C3H8) react with excess oxygen

Answer 1

In this problem the limiting reactant is not oxygen because you have been given it in excess. So by default propane is the limiting reactant (C3H8).

You will now use the limiting reactant to solve the problem.

You will make sure the equation is balanced on both sides.

Then you will use stoichiometry to determine how many moles can be produced of the H2O(g) on the product side.

This equation:

C3H8(g) + 5O2(g) ---> 3CO2(s) + 4H2O(g)

is a balanced equation.

You have 3 Carbons on each side

8 Hydrogens on each side

and

10 Oxygens on each side

So going by the coeffecients in front of each species of reactants and products:

1 mole of propane (C3H8) yields 4 moles of H20

So if 7.9 moles of propane (C3H8), how many moles of H20 would that yield?

set up a ratio and solve for X:

1 mol (C3H8) = 4 mol (H2O)

_______________________

7.9 mol (C3H8)= X mol (H2O)

7.9*(4) = 1*X

31.6 = X

So 7.9 moles of propane (C3H8) would yield 31.6 moles of water (H2O), you could round to 32 moles of H2O for sig. fig purposes.