Question

Identify the coefficient $x,y,z$ respectively in the following reaction:
$xMnO+yPb{O}_2+zHN{O}_3\to HMn{O}_4+Pb(N{O}_3{)}_2+H_{2}O$

• A. 2, 5, 10
• B. 5, 2, 10
• C. 10, 5, 2
• D. 2, 5, 4

Answer 1

The correct option is A 2, 5, 10

Given reaction
$MnO+Pb{O}_2+HN{O}_3\to HMn{O}_4+Pb(N{O}_3{)}_2$

Oxidation half reaction:
$MnO\to HMn{O}_4$
Step 1 - Balance atoms other than $O$ and $H$
$MnO\to HMn{O}_4$
Step 2 - Balance O atoms from $H_{2}O$
$MnO+3H_{2}O\to HMn{O}_4$
Step 3 - Balance H atoms from $H^{+}$
$MnO+3H_{2}O\to HMn{O}_4+5H^{+}$
Step 4 - Balance charge
$MnO+3H_{2}O\to HMn{O}_4+5H^{+}+5e^{-}..\left(1\right)$

Similiarly, reduction half reaction:
$Pb{O}_2+HN{O}_3\to Pb(N{O}_3{)}_2$
$Pb{O}_2+2HN{O}_3\to Pb(N{O}_3{)}_2$
$Pb{O}_2+2HN{O}_3\to Pb(N{O}_3{)}_2+2H_{2}O$
$Pb{O}_2+2HN{O}_3+2H^{+}\to Pb(N{O}_3{)}_2+2H_{2}O$
$Pb{O}_2+2HN{O}_3+2H^{+}+2e^{-}\to Pb(N{O}_3{)}_2+2H_{2}O...(2)$

Now, Multiply Eq. (1) by 2 and (2) by 5
and then add we have
$2MnO+5Pb{O}_2+10HN{O}_3\to 2HMn{O}_4+5Pb(N{O}_3{)}_2+4H_{2}O$

This is the balanced chemical Equation.
$\therefore$ $x,y$ and $z$ are 2, 5 and 10.