Question

How many grams of water can be produced when 11.7 moles of ethane (C2H6) react with excess oxygen gas?

Answer 1

Answer: 631.8 g

Explanation:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

It can be seen from the balanced chemical equation, 2 moles of ethane reacts with 7 moles of Oxygen gas to produce 4 moles of carbon dioxide and 6 moles of water.

Ethane is the limiting reagent as it limits the formation of product.

Thus, if  2 moles of ethane produce 6 moles of water.

11.7 moles moles of ethane produce=$\frac{6}{2}\times 11.7=35.1 moles$of water.

Mass of water= no of moles$\times$ Molar mass

Mass of water= 35.1$\times$ 18g/mol= 631.8 g

Answer 2

SO the way we balance this is like this:
2C2H6 + 7O2 → 4CO2 + 6H2O
x/11.7 = 6/2 x = 35.1 moles of of water are produced
35.1 mol * 18 g/mol = 631.8 g of water are produced.
Please take into account that: 
2 moles of ethane gives =6 moles of water
11.7 moles of ethane gives = X moles of water (assume) cross multiplication =(6*11.7)/2
=35.1 moles
I hope this can help you greatly