Question

From a laboratory process, a student collects 28.0 g of hydrogen and 224.0 g of oxygen. how much water was originally involved in the process?

Answer 1

Chemical reaction is given as :

`2H_(2)O\rightarrow 2H_(2)+O_(2)`

Here, 2 moles of water gives 2 moles of hydrogen and one mole of oxygen.

Mass of hydrogen = 28.0 g (given)

Mass of oxygen = 224.0 g (given)

Number of moles =
`(given mass)/(molar mass)`

Thus, number of moles of hydrogen =
`(28 g)/(2* 1 g/mol)`

= 14 moles of hydrogen

Number of moles of oxygen=
`(224 g)/(2* 16 g/mol)`

= 7 moles of oxygen

Now, to find the amount of water:

`7 mol of oxygen * ((2 mol of water)/(1 mol of oxygen))* (18 g/mol of water)`

= 252 g of water.

Thus, amount of water is 252 g (involved in the process).