How many grams of water can be produced when 11.7 moles of ethane (c2h6) react with excess oxygen gas?

unbalanced equation: c2h6 + o2 → co2 + h2o

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

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