How many grams of oxgyen `(O_(2))` are required to completely react with 0.300 gof hydrogen `(H_(2))` to yield `H_(2)O`? Also calculate the amount of water formed.
`2H_(2) + O_(2) to 2H_(2)O`

Explanation

`underset(4g)underset(2 mol)(2H_(2)) + underset(32 g)underset(1 mol)(O_(2)) to underset(36 g)underset(2mol)(2H_(2)O)`
Now, 4g of `H_(2)` require oxygen = 32 g
0.300 g of `H_(2)` require oxygen `=32/4 xx 0.300 = 2.4 `g
Again, 42of `H_(2)` produce `H_(2)O = 36 g`
0.300 g of `H_(2)` produces `H_(2)O = 36/4 xx 0.300 = 2.7`g