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

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)

6 CO2 + 6 H2O + energy -------> C6H12O6 + 6 O2

So, 6 moles of CO2 required for 1 mole of glucose production.

i.e, for 1mole of glucose (= 180 grams) it needs 44*6= 264 g of CO2.

therefore for 20 g of glucose, 264/9= 29.33 g of CO2 required.

Please tell me this is a troll question.

1ml is equal to 1g. This only applies to pure water.

Other liquids have different densities.

So 150 ml of water= 150 grams of water.

That's all from me….

Hope it may be helpful….