If 30 Grams Of Nonane, C9H20, Undergoes Combustion, How Many ...

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Downvote + Chemical reactions + Water + Chemistry + Organic chemistry Posted by Kim Aaron If 30 grams of nonane, C9H20, undergoes combustion, how many grams of water would be produced?

C9H20 + 14O2 = 9CO2 + 10H2O

1 mole nonane +14 moles O2 = 9 moles CO2 + 10 moles H2O

30 gr of nonane = how many moles? X

X mole os nonane will produce how many moles H2O? Y

Then convert Y into grams

David Shanen Follow Following

C9H20 + 14O2 = 9CO2 + 10H2O

1 mole nonane +14 moles O2 = 9 moles CO2 + 10 moles H2O

30 gr of nonane = how many moles? X

X mole os nonane will produce how many moles H2O? Y

Then convert Y into grams

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Balanced equation:

C9H20 + 19O2 -> 9CO2 + 10H2O

Molar mass of nonane = (12 × 9) + (1 × 20)

= 128 g/mol

Moles in 30 g of nonane = 30/128 = 0.234 mol

Stoichiometrically, 1 mole of nonane produces 10 moles of water.

Moles of water produced by 0.234 mol of nonane = 0.234 × 10 = 2.34 moles

Molar mass water = 18 g/mol

Mass of 2.34 mol of water = 18 g × 2.34

= 42.1 g (answer)

Christine EveningsFall Follow Following

Balanced equation:

C9H20 + 19O2 -> 9CO2 + 10H2O

Molar mass of nonane = (12 × 9) + (1 × 20)

= 128 g/mol

Moles in 30 g of nonane = 30/128 = 0.234 mol

Stoichiometrically, 1 mole of nonane produces 10 moles of water.

Moles of water produced by 0.234 mol of nonane = 0.234 × 10 = 2.34 moles

Molar mass water = 18 g/mol

Mass of 2.34 mol of water = 18 g × 2.34

= 42.1 g (answer)

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