How would I solve this problem?

Question

Ethanol (C2H5OH) is synthesized for industrial use by the following reaction, carried out at very high pressure.
C2H4(g) + H2O(g) C2H5OH(l)
What is the maximum amount of ethanol that can be produced when 1.3 kg of ethylene (C2H4) and 0.013 kg of steam are placed into the reaction vessel?

I am confused as to how I answer this question. Can someone show me how to work it out?

Answer 1

The equation is balanced: 1 mole of C2H4 reacts with 1 mole of H2O for form 1 mole of C2H5OH.
1,300 g of ethylene is how many moles?
13 grams of steam (H2O) is how many moles?
Which reagent is limiting and which is present in excess?
The moles of the limiting reagent will yield the same number of moles of ethanol.
Convert the moles of ethanol to grams of ethanol.

You can see immediately that H2O (steam) is the limiting reagent. You have only 0.72 moles of H2O (13g / 18 g/mole).
You have 46+ moles of ethylene (1300g / 28g/mole).

Thwe maximum amount of ethanol you can get is 0.72 moles or 33 gram (0.72 moles x 46 grams/mole)

Answer 2

The equation seems balances, so we move on.
Either ethylene or water will be a limiting reactant. The reaction indicates they react in a 1:1 ratio.
So divide each mass term by mole weight to get moles. The one with the least moles is limiting, and the ethanol yield is limited to that number of moles (mult. by mole wt to get kgs).