Question

Balance the following chemical equations:
$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+NO+H_{2}O$

Answer 1

$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+NO+H_{2}O$

Let us first balance, ${}^{\prime }{H}^{\prime }$ and ${}^{\prime }{O}^{\prime }$ on both sides of the equation, by inserting ${}^{\prime }{8}^{\prime }$ before $HN{O}_3$ in L.H.S.; $3$ before $Cu(N{O}_3{)}_2$; $2$ before $NO$ and $4$ before $H_{2}O$ in R.H.S., respectively, we get:"

$Cu+8HN{O}_3\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

Now, to balance ${}^{\prime }C{u}^{\prime }$ on both sides, insert ${}^{\prime }{3}^{\prime }$ before $Cu$ in L.H.S., we get:

$3Cu+8HN{O}_3\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

This is balanced equation.