Question

$6$ g of $H_2$ reacts with $14$ g of $N_2$ to form $N{H}_3$ till the reaction completely consumes the limiting reagent. The amount of another reactant in $g$ left is:

Answer 1

The ba;lanced reaction is as follows:
$N_2+3H_2\to 2N{H}_3$
$28$ g $6$ g $34$ g
$14$ g $6$ g (Given)
$14$ gm $3$ gm (Actually consumed)
Here, nitrogen is limiting reagent so, excess hydrogen left is $3$ g.