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 = $\frac{given mass}{molar mass}$

Thus, number of moles of hydrogen = $\frac{28 g}{2\times 1 g/mol}$

= 14 moles of hydrogen

Number of moles of oxygen= $\frac{224 g}{2\times 16 g/mol}$

= 7 moles of oxygen

Now, to find the amount of water:

$7 mol of oxygen \times (\frac{2 mol of water}{1 mol of oxygen})\times (18 g/mol of water)$

= 252 g of water.

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