The combustion of a sample of butane, C4H10 (lighter fluid), produced 210.46 grams of water.im trying to find the molar mass of water in this equation: 2C4H10+13O2----8CO2+10H2O

Question

The combustion of a sample of butane, C4H10 (lighter fluid), produced 210.46 grams of water.im trying to find the molar mass of water in this equation: 2C4H10+13O2---->8CO2+10H2O

anonymous · Answer

do i have to multiply 210.26g of h2o by 1 mole of h2o or by 10 moles of h2o as it says in the equation

koikkara · Answer

Hai Sweet Friend, @nicholas127  WelcoMe To OpeN studY !!

>>By 10 moles of h2o as it says in the equation !

Answer Hint:
Molar mass of butane (C4H10) is 58.14 g/mol.
So, 5.90 g / 58.14 g/mol = .10148 mol
Using mole ratios : .10148 mol C4H10 * (10 mol H2O / 2 mol C4H10) = .5074 mol H2O
Molar mass of H2O: 18.02 g/mol
Thus, .5074 mol * 18.02 g/mol = 9.1434 g H2O 

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