How much water would be formed in 4.8 grams of hydrogen reacted with 38.4 grams of oxygen?

Question

Please give a detailed example on how to solve it. Math and reasons on how you solved it, thanks.

Answer 1

First determine the limiting reagent (the one that runs out first) because that will determine the maximum amount of product could possibly form.

there are 4.8 moles of hydrogen [4.8g/1g/mol]

there are 2.4 moles of oxygen 38.4g/16g/mol

Since the formula for water is H2O we need twice as much hydrogen as oxygen, so in our case, there is no limiting reagent, everything gets used up.

The 2.4 moles of oxygen will yield 2.4 moles of water as one molecule of oxygen will yield one molecule of water

[2.4 x (16g + [2x1g])] or 43.2g of water.

Answer 2

Convert grams to moles.

grams x Mol. Wt. = moles

For hydrogen mol wt is 1.0 g/mol therefore:

4.8 g / 1 g/mol = 4.8 mol hydrogen

For oxygen mol wt is 16 g/mol therefore:

38.4 g / 16 g/mol = 2.4 mol oxygen

Water is H2O; it contains two atoms hydrogen and one atom oxygen. Hydrogen is H2 (diatomic) so one molecule of hydrogen contains TWO hydrogen atoms. So is oxygen. So TWO moles of hydrogen must react with one mole oxygen to make water.

2H2 + O2 = 2H2O
Start 4.8 2.4 0
End 0 0 2.4

Conveniently, you have exactly double the number of moles of hydrogen than you do oxygen, so you have no limiting reagent to worry about. You have 2.4 mol of oxygen, so you will end up with 2.4 mol of water.

The molecular weight of water is 18 g/mol, so convert your 2.4 mol of water back to grams.

2.4 mol x 18 g/mol = 43.2 g water

But, of course, water is a liquid. You measure liquids in millilitres. Fortunately, we know that the density of water is 1 g/mL, so just convert the grams to millilitres.

43.2 g / 1 g/mL = 43.2 mL water

43.2 mL of water is produced.

Answer 3

7.2 grams of pure water