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How many grams of water are produced by the combustion of 32.0g of ch4?

CH4+202–>2H2+CO2
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To determine the amount of water produced by the combustion of 32.0g of CH4, we need to first calculate the amount of CH4 that reacts and the amount of H2O that is produced.

The balanced chemical equation for the combustion of CH4 is:

CH4 + 2O2 → CO2 + 2H2O

From this equation, we see that for every 1 mole of CH4 that reacts, 2 moles of H2O are produced.

To calculate the amount of CH4 that reacts, we need to convert the given mass of CH4 into moles using its molar mass:

32.0 g CH4 × (1 mol CH4/16.04 g CH4) = 1.996 mol CH4

Therefore, 1.996 moles of CH4 react.

Since 2 moles of H2O are produced for every 1 mole of CH4 that reacts, we can calculate the amount of H2O produced:

2 mol H2O/mol CH4 × 1.996 mol CH4 = 3.992 mol H2O

Finally, we can convert the amount of H2O produced in moles to grams using its molar mass:

3.992 mol H2O × (18.02 g H2O/1 mol H2O) = 72.1 g H2O

Therefore, the combustion of 32.0g of CH4 produces 72.1 grams of H2O
User Sangeet Shah
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