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.

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.

Owen Ambrose · Accepted Answer

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"#.

Naomi Aronson · Answer

#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#

Ellie Andrews · Answer

undefined To find the mass of water formed in the complete combustion of 5.00 x 10^3 g of CH4, we first calculate the moles of CH4 using its molar mass, which is 16.04 g/mol. Then, using the stoichiometry of the reaction, we determine the moles of water formed and convert it to grams. Finally, we convert grams to kilograms.

Molar mass of CH4 = 16.04 g/mol
5.00 x 10^3 g CH4 / 16.04 g/mol = 311.22 mol CH4
From the balanced chemical equation:
1 mole of CH4 produces 2 moles of H2O
So, 311.22 mol CH4 * 2 mol H2O / 1 mol CH4 = 622.44 mol H2O
Molar mass of H2O = 18.02 g/mol
622.44 mol H2O * 18.02 g/mol = 11,217.94 g H2O

Converting grams to kilograms:
11,217.94 g H2O / 1000 = 11.2 kg

Therefore, the mass of water formed in the complete combustion of 5.00 x 10^3 g of CH4 is 11.2 kg.

Caleb Adams · Answer

undefined The molar mass of water (H2O) is 18.015 g/mol. In the balanced chemical equation for the combustion of methane (CH4) to form water (H2O), the stoichiometric coefficient of water is 2. This means that 2 moles of water are produced for every 1 mole of methane consumed.

First, we need to find the number of moles of methane in 5.00 x 10^3 g:
$ n_{\text{CH}_4} = \frac{5.00 \times 10^3 \text{ g}}{16.04 \text{ g/mol}} = 311.8 \text{ mol} $

Next, we determine the number of moles of water produced using the stoichiometric ratio:
$ n_{\text{H}_2\text{O}} = 2 \times n_{\text{CH}_4} = 2 \times 311.8 \text{ mol} = 623.6 \text{ mol} $

Now, we find the mass of water produced:
$ m_{\text{H}_2\text{O}} = n_{\text{H}_2\text{O}} \times \text{molar mass of water} = 623.6 \text{ mol} \times 18.015 \text{ g/mol} = 11,222.4 \text{ g} $

Converting grams to kilograms:
$ 11,222.4 \text{ g} = 11.2 \text{ kg} $

Therefore, the mass of water formed in the complete combustion of 5.00 x 10^3 g of CH4 is 11.2 kg.