A sample of 10gm of `H_(2)` reacts with sufficient amount of oxygen to form 106gm of `H_(2)O(l)` and `H_(2)O_(2)(l)`. Calculate mass % of `H_(2)O`.

Options

A. 0.679
B. 0.321
C. 0.152
D. 0.848

Explanation

Assume a mole `H_(2)O`
b mole `H_(2)O_(2)`
`axx18+bxx34=106` . . . (i)
`POAc` on H:
`2xxn_(H_(2))=2xxn_(H_(2)O)+2xxn_(H_(2)O_(2))`
`2xx(10)/(2)=10=2a+2b`
`a+b=5` . . . .(ii)
`18a+18b=90`. . . . . . ..(ii)
`18a+18b=106` . . . . . . .(i)
16b=16
b=1
a=4
Mass of `H_(2)O=axx18`
`=4xx19=72gm`
Mass `% H_(2)O=(72)/(106)xx100=67.9%`