In the redox reactionx MnO + y PbO_2 + z H | vedclass

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(A) The redox reaction is: $x \ MnO + y \ PbO_2 + z \ HNO_3 	o a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$.Step $1$: Assign oxidation states. $Mn$ in $Mn

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$x = 2, y = 5, z = 10$

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(A) The redox reaction is: $x \ MnO + y \ PbO_2 + z \ HNO_3 	o a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$.Step $1$: Assign oxidation states. $Mn$ in $MnO$ is $+2$,$Mn$ in $HMnO_4$ is $+7$ (change of $+5$). $Pb$ in $PbO_2$ is $+4$,$Pb$ in $Pb(NO_3)_2$ is $+2$ (change of $-2$).Step $2$: Balance the change in oxidation states. Multiply $MnO$ by $2$ and $PbO_2$ by $5$ to balance the electrons ($2 	imes 5 = 10$ and $5 	imes 2 = 10$).This gives: $2 \ MnO + 5 \ PbO_2 + z \ HNO_3 	o 2 \ HMnO_4 + 5 \ Pb(NO_3)_2 + c \ H_2O$.Step $3$: Balance Nitrogen atoms. There are $5 	imes 2 = 10$ nitrogen atoms on the right,so $z = 10$.Step $4$: Balance Hydrogen atoms. There are $10$ hydrogen atoms on the left,so $c = 4$ to give $4 \ H_2O$.The balanced equation is: $2 \ MnO + 5 \ PbO_2 + 10 \ HNO_3 	o 2 \ HMnO_4 + 5 \ Pb(NO_3)_2 + 4 \ H_2O$.

In the redox reaction
$x \ MnO + y \ PbO_2 + z \ HNO_3 \to a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$
Find the values of $x, y, z$.

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In the redox reaction$x \ MnO + y \ PbO_2 + z \ HNO_3 \to a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$Find the values of $x, y, z$.