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.

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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.

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Answer 1 · 2015-02-21T16:36:14

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

Start with the balanced chemical equation

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ $CH_4 + 2O_2 -> CO_2 + 2H_2O$

Now look at the mole ratio that exists between $C H_{4}$ $CH_4$ and $H_{2} O$ $H_2O$ : 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 $C H_{4}$ $CH_4$ react, assuming that oxygen is not a limiting reagent.

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

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

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

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

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

Expressed in kilograms, this is equal to

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

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

Answer 2 · 2015-02-21T16:39:20

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ $CH_4+2O_2rarrCO_2+2H_2O$

Imole $\to$ $rarr$ 2 moles

Convert to grams.

( $A_{r} C = 12, A_{r} H = 1, A_{r} O = 16)$ $A_rC=12,A_rH=1,A_rO=16)$

$[12 + (4 \times 1)] g \to 2 \times [16 + (2 \times 1)] g$ $[12+(4xx1)]grarr2xx[16+(2xx1)]g$

$16 g \to 2 \times 18 = 36 g$ $16grarr2xx18=36g$

I have assumed the mass of methane to be $5 \times 10^{3} g$ $5xx10^3g$ .

So $1 g \to \frac{36}{16} g$ $1grarr36/16g$

So $5 \times 10^{3} g \to \frac{36}{16} \times 5 \times 10^{3} g$ $5xx10^(3)grarr(36)/(16)xx5xx10^(3)g$

$= 11.25 \times 10^{3} g$ $=11.25xx10^(3)g$

$= 11.25 k g$ $=11.25kg$

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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.

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