Given this reaction: $CH_4 + 2O_2 -> CO_2 + H_2O$ , 16 g of $CH_4$ react with 64 g of $O_2$ , producing 44 g of $CO_2$ . How many grams of water are produced?

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Answer 1 · 2016-11-29T14:20:18

$36$ grams of water will get formed.

There are actually two methods of doing it - stoichimetry & also with the help of law of conservation of mass.

First by stoichiometry:

$16$ grams of $CH_4$ means $1$ mole of it.

Now, after balancing the entire equation we get:

$CH_4+2O_2 rarr CO_2+2H_2O$ .

So, 1 mole of methane produces 2 moles of water. Hence, $36$ grams of water is formed.

Now by law of conservation of mass:

The entire mass of the reactants will be equal to the entire mass of the products after the reaction.

So, mass of water formed $=(16+64)-44$ grams, i.e. $36$ grams.

Given this reaction: CH_4 + 2O_2 -> CO_2 + H_2OCH_4 + 2O_2 -> CO_2 + H_2O, 16 g of CH_4CH_4 react with 64 g of O_2O_2, producing 44 g of CO_2CO_2. How many grams of water are produced?