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

Options

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

Answer

C

Explanation

`underset(2mol)(2H_(2)(g)) + underset(1mol)(O_(2)(g)) rarr underset(2mol)(2H_(2) O(l))`
`n_(H_(2)) = (10g)/(2g mol^(-1)) = 5 mol`
`n_(O_(3)) = (64g)/(32g mol^(-1)) = 2 mol`
Moles of product formed from each rectant:
`5 mol H_(2) xx (2 mol H_(2) O)/(2 mol H_(2)) = 5 mol H_(2) O`
`2 mol O_(2) xx (2 mol H_(2)O)/(1 mol O_(2) = 4 mol H_(2) O`
Thus, `O_(2)` is the limiting reacant and `4 mol H_(2) O` will be produced.