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Consider the following reaction , N(2) +...

Consider the following reaction , `N_(2) + 3H_(2) to 2NH_(3)`
Molecular weight of `NH_(3)` and `N_2` are `x_(1)" and "x_(2)` respectively. Their equivalent weights are `y_(1)" and "y_(2)` respectively. Then `(y_(1)-y_(2))` is

A

`(( 2 x _(1) - x _(2))/( 6))`

B

`(x _(1) -x _(2))`

C

`(3x _(1) -x _(2))`

D

`(x _(1) - 3x _(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`N _(2) = 2 NH _(3) = 3 H _(2) = 6H`
eq. mass of `N _(2) = (x _(2))/(6) = y _(2)` eq. mass of `NH _(3) = ( 2 x _(1))/(6) = y _(4)`
1 eq. of `N _(2)` combines with 1 eq. of `H _(2)` of form 1 eq. of `NH _(3)`
`therefore (y _(1) - y _(2)) = ((2 x _(1) )/(6) - (x _(2))/( 6)) = ((2 x _(1) - x _(2))/( 6)) `
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