Methane, CH4 , the major component of natural gas burns in air to form CO2 and H2O. What mass of water is formed in the complete combustion of 5.00 x 103 g of CH4? Why is the answer: 11.2 kg. Thank you.

Yes, the answer is approximately #"11.2 kg"#, but actually closer to #"11.3 kg"#

Start with the balanced chemical equation

#CH_4 + 2O_2 -> CO_2 + 2H_2O#

Now look at the mole ratio that exists between #CH_4# and #H_2O#: 1 mole of methane will produce 2 moles of water.

In order to see how many moles of water are produced, you must determine how many moles of #CH_4# react, assuming that oxygen is not a limiting reagent.

#5.00 * 10^(3) "g methane" * ("1 mole")/("16.0 g") = "312.5 moles methane"#

According to the aforementioned mole ratio, the number of moles of water produced will be

#"312.5 moles methane" * ("2 moles water")/("1 mole methane") = "625 moles water"#

Now just use water's molar mass to calculate the actual mass produced

#"625 moles" * ("18.0 g")/("1 mole") = "11,250 g"#

Expressed in kilograms, this is equal to

#"11,250 g" * ("1 kg")/("1000 g") = "11.25 kg of water"#

If you round this to three sig figs, you'll actually get #"11.3 kg"#.

#CH_4+2O_2rarrCO_2+2H_2O#

Imole #rarr#2 moles

Convert to grams.

( #A_rC=12,A_rH=1,A_rO=16)#

#[12+(4xx1)]grarr2xx[16+(2xx1)]g#

#16grarr2xx18=36g#

I have assumed the mass of methane to be #5xx10^3g#.

So #1grarr36/16g#

So #5xx10^(3)grarr(36)/(16)xx5xx10^(3)g#

#=11.25xx10^(3)g#

#=11.25kg#