$h_2 + n_2 \ ightarrow nh_3$
$3\text{H}_2 + \text{N}_2 ightarrow 2\text{NH}_3$
$3\text{H}_2 + \text{N}_2 ightarrow 2\text{NH}_3$
$h_2 + n_2 \
ightarrow nh_3$
$h_2 + n_2 \
ightarrow nh_3$
The given unbalanced reaction is:
$\text{H}_2 + \text{N}_2
ightarrow \text{NH}_3$
There are 2 N on left, add coefficient 2 to $\text{NH}_3$:
$\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
There are 6 H on right, add coefficient 3 to $\text{H}_2$:
$3\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
$3\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
The given unbalanced reaction is:
$\text{H}_2 + \text{N}_2
ightarrow \text{NH}_3$
There are 2 N on left, add coefficient 2 to $\text{NH}_3$:
$\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
There are 6 H on right, add coefficient 3 to $\text{H}_2$:
$3\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
$3\text{H}_2 + \text{N}_2
ightarrow 2\text{NH}_3$
$h_2 + n_2 \ ightarrow nh_3$
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x equals 14
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