When a 1.52 ng sample of methane burns in excess oxygen, how many grams of water and carbon dioxide are formed
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When a 1.52 ng sample of methane burns in excess oxygen, how many grams of water and carbon dioxide are formed

[From: ] [author: ] [Date: 12-04-14] [Hit: ]
1.52 ng * 44/16 = 4.1.52 ng * 36/16 = 3.......
I keep getting around 0.00000000003 and the answer is wrong, but I know I wrote the conversion path right..

1.52 ng x 1x10^-9 g / 1 ng x 1 mol CH4/ 16.05g x 2 mol H2O / 1 CH4 x 18.02g/ 1mol H20

Can someone find my mistake? Which I doubt there is one.

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First, balance the equation

CH4 + 2O2 --> CO2 + 2H2O

Next 1.52 ng of CH4 is how many moles? Divide by 16 g/mole. This gives approximately 10^-10 moles of CH4 which produce exactly that many moles of CO2 and twice that many moles of H2O.

CO2 has a molecular mass of 44 g/mole, so you get about 4.4 x 10^-9 g of CO2
H2O has a molecular mass of 18 g/mole so you get about 3.5 x 10^-9 g of H2O

My math is being done sans calculator and you will need to report the values to three significant figures, Which will be something like 0.0000000044 for CO2

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CH4 weighs 16
and CO2 weighs 44
2 H2O weighs 36


1.52 ng * 44/16 = 4.18 ng (for CO2)
1.52 ng * 36/16 = 3.42 ng (for water)

Soc dont need no stickn moles to solve these
1
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