# 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.

Feb 21, 2015

Yes, the answer is approximately $\text{11.2 kg}$, but actually closer to $\text{11.3 kg}$

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$

Now look at the mole ratio that exists between $C {H}_{4}$ and ${H}_{2} O$: 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 $C {H}_{4}$ react, assuming that oxygen is not a limiting reagent.

$5.00 \cdot {10}^{3} \text{g methane" * ("1 mole")/("16.0 g") = "312.5 moles methane}$

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

$\text{312.5 moles methane" * ("2 moles water")/("1 mole methane") = "625 moles water}$

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

$\text{625 moles" * ("18.0 g")/("1 mole") = "11,250 g}$

Expressed in kilograms, this is equal to

$\text{11,250 g" * ("1 kg")/("1000 g") = "11.25 kg of water}$

If you round this to three sig figs, you'll actually get $\text{11.3 kg}$.

Feb 21, 2015

$C {H}_{4} + 2 {O}_{2} \rightarrow C {O}_{2} + 2 {H}_{2} O$

Imole $\rightarrow$2 moles

Convert to grams.

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

$\left[12 + \left(4 \times 1\right)\right] g \rightarrow 2 \times \left[16 + \left(2 \times 1\right)\right] g$

$16 g \rightarrow 2 \times 18 = 36 g$

I have assumed the mass of methane to be $5 \times {10}^{3} g$.

So $1 g \rightarrow \frac{36}{16} g$

So $5 \times {10}^{3} g \rightarrow \frac{36}{16} \times 5 \times {10}^{3} g$

$= 11.25 \times {10}^{3} g$

$= 11.25 k g$