10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

Options

A. 1 mol
B. 2 mol
C. 3 mol
D. 4 mol

Answer

D

Explanation

`underset(2g)(H_(2))+underset(16g)(1//2O_(2))to underset(18g)(H_(2)O)`
10g `H_(2)` required `O_(2)=80` which is not present 64 g `O_(2)` requird 8g of `H_(2) and H_(2) ` left =2g . Thus , `O_(2)` is the limiting rectant and `H_(2)` is excess reactant .
Hence , `H_(2)O` formed from 64 of `O_(2)`
`=(18)/(16)xx 64`
` =72 g= (72)/(18) ` mole
= 4 mole