2) C_H,(g) + 30,(g) → 2 CO2(g) + 2 H2O(g)
a) 2.7 mol of C, H, is reacted with 6.3 mol of 0,, how many moles of water will be
made?
b) What is the limiting reactant?
c) What is the excess reactant?

a. 2.7 mol of water

b CH2.

c. O2

Explanation:

The complete equation of the reaction should be:

2CH2(g) + 3O2(g) → 2 CO2(g) + 2 H2O(g)

a) how many moles of water will be  made?

To make 2 molecules of water (H2O) we need 2 molecules of CH and 3 molecules of O2.

We have 2.7 mol of CH2, the possible yield of water produced if it all used up will be:

2.7 mol * 2/2= 2.7 mol

We have 6.3 mol of O2, the possible yield of water produced  if it all used up will be:

6.3 mol * 2/3 = 4.2 mol

Since the maximum yield of CH2 lower, we can have 2.7 mol of water and have some excess oxygen at the end of the reaction.

b) What is the limiting reactant?

A limiting reactant is a reactant that will be used up in the reaction. This reactant has the lowest stoichiometric ratio compared to other reactants, which make them the one depleted out first. Since they depleted, the reaction will stop. Thus they limit the number of reactions and called limiting reactants. If you add the limiting reactant, the reaction will continue.

The limiting reactant in this reaction is the CH2. When producing water molecules, all 2.7 mol of CH2 will be used while we still have O2 left.

c) What is the excess reactant?

The excess reactant will have some remains after the reaction stop. That is because the excess reactant has more mass than needed for the reaction that will use all limiting reactants. Since we still have remains, adding excess reactant won't continue the reaction.

The excess reactant in this question is O2 since it still has remained after we make 2.7 mol of water. The O2 remaining, in this case, will be:

6.3 mol - 2.7mol * 3/2= 2.25 moles

it's filtration, but this is an assumption so my apologies if i got this