Question

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.

Answer 1

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"`.

Answer 2

`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`