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How many grams of water are produced when 35.0 grams of methane (CHâ‚„) reacts with oxygen (Oâ‚‚) in the following balanced equation?

\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

A. 4.38 g
B. 78.8 g
C. 126 g
D. 39.4 g

Answer :

Final answer:

The reaction of 35.0 grams of methane with oxygen produces 78.8 grams of water, according to the stoichiometry of the balanced chemical equation.

Explanation:

The question is about a chemical reaction where methane (CHâ‚„) reacts with oxygen (Oâ‚‚) to produce carbon dioxide (COâ‚‚) and water (Hâ‚‚O). To find how many grams of water is produced from 35.0 grams of methane you can use the stoichiometry of the balanced chemical equation.

Firstly, we need to calculate the mole of methane.The molar mass of Methane (CH4) is approximately 16.04 g/mol. Therefore, the number of moles of methane is 35 g / 16.04 g/mol = 2.18 moles. From the balanced equation, it is known that 1 mole of CHâ‚„ produces 2 moles of Hâ‚‚O. Thus, 2.18 moles of CHâ‚„ will produce 2*2.18 = 4.36 moles of Hâ‚‚O.

Secondly, turning moles of Hâ‚‚O to grams. The molar mass of water (H2O) is approximately 18.02 g/mol. Therefore, 4.36 moles of H2O will weigh 4.36 moles * 18.02 g/mol = 78.6 g.

So, the answer is b. 78.8 g

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Rewritten by : Jeany