How many gram of oxygen `(O_(2))` is required to completely react with 0.200 g of hydrogen `(H_(2))` to yield water `(H_(2)O)` ? Also calculate the amount of water formed (molecular mass H = 2, O = 32).

Explanation

The balanced equation for the reaction is
`2H_(2)+ O_(2) to 2H_(2)`
Now, 4 g of H, require oxygen = 32 g
0.200 g of `H_(2)` require oxygen `= (32)/(4) xx 0.200= 1.6 g`
Again, 4 g of `H_(2)` produce `H_(2)O` = 36 g.
0.200 g of `H_(2)` produce `H_(2)O= (36)/(4) xx 0.200 = 1.8g`