Question

Balance the following equations.
1.NaNO3+PbO-------------Pb9NO302+Na2O
2.Agl+Fe2(CO3)3-------------Fel3+AgCO3
3.C2H4O2+O2-------------CO2+H2O
4.ZnSO4+Li2CO3----------------ZNCO3+Li2SO4
5.V2O5+CaS----------------CaO+V2S5

Answer 1

Answer:

Maybe helpful.

Answer 2

Answer:

The balanced chemical equations are,

$1\ )$ $2\ NaNO_{3} \ +\ PbO\ --->\ Pb(NO_{3})_2\ +\ Na_{2}O$

$2\ )$ $6\ AgI \ +\ Fe_2(CO_3)_3\ --->\ 2\ FeI_{3}\ +\ 3\ Ag_{2}CO_3$

$3\ )$ $C_2H_4O_2\ +\ 2\ O_2\ --->\ 2\ CO_2\ +\ 2\ H_2O$

$4\ )$ $ZnSO_4\ +\ Li_2CO_3\ --->\ ZnCO_3\ +\ Li_2SO_4$

$5\ )$ $V_2O_5\ +\ 5\ CaS\ --->\ 5\ CaO\ +\ V_2S_5$

Explanation:

To balance a chemical equation,

First, we need to count the atoms in both the reactant and the product sides· Then, add coefficients suitable for that·

$1\ )$  $NaNO_{3} \ +\ PbO\ --->\ Pb(NO_{3})_2\ +\ Na_{2}O$

Here,

No· of· atoms on the reactant side is,

$Na\ =\ 1\ ,\ \ NO_3\ =\ 1\ ,\ \ Pb\ =\ 1\ ,\ \ O\ =\ 1$

No· of· atoms on the product side is,  

$Na\ =\ 2\ ,\ \ NO_3\ =\ 2\ ,\ \ Pb\ =\ 1\ ,\ \ O\ =\ 1$

To balance this equation, add $2$ near the $NaNO_{3}$ on the reactant side·

So, we get the balanced chemical reaction as,

$2\ NaNO_{3} \ +\ PbO\ --->\ Pb(NO_{3})_2\ +\ Na_{2}O$

$2\ )$  $AgI \ +\ Fe_2(CO_3)_3\ --->\ FeI_{3}\ +\ Ag_{2}CO_3$

The correct chemical formula for the silver carbonate is $Ag_2CO_3.$

Here,

No· of· atoms on the reactant side is,

$Ag\ =\ 1\ ,\ \ I\ =\ 1\ ,\ \ Fe\ =\ 2\ ,\ \ CO_3\ =\ 3$

No· of· atoms on the product side is,

$Ag\ =\ 2\ ,\ \ I\ =\ 3\ ,\ \ Fe\ =\ 1\ ,\ \ CO_3\ =\ 1$

To balance this equation, add $3$ near the $Ag_2CO_3$ in the product side and then add $2$ near the $FeI_3$ and $6$ near the $AgI$ following that·

So, we get the balanced chemical reaction as,

$6\ AgI \ +\ Fe_2(CO_3)_3\ --->\ 2\ FeI_{3}\ +\ 3\ Ag_{2}CO_3$

$3\ )$ $C_2H_4O_2\ +\ O_2\ --->\ CO_2\ +\ H_2O$

Here,

No· of· atoms on the reactant side is,

$C\ =\ 2\ ,\ \ H\ =\ 4\ ,\ \ O\ =\ 4$

No· of· atoms on the product side is,

$C\ =\ 1\ ,\ \ H\ =\ 2\ ,\ \ O\ =\ 3$

To balance this equation, add $2$ near the $CO_2$ in the product side and then add $2$ near the $H_2O$ and $2$ near the $O_2$  following that·

So, we get the balanced chemical reaction as,

$C_2H_4O_2\ +\ 2\ O_2\ --->\ 2\ CO_2\ +\ 2\ H_2O$

$4\ )$ $ZnSO_4\ +\ Li_2CO_3\ --->\ ZnCO_3\ +\ Li_2SO_4$

Here,

No· of· atoms on the reactant side is,

$Zn\ =\ 1\ ,\ \ SO_4\ =\ 1\ ,\ \ Li\ =\ 2\ ,\ \ CO_3\ =\ 1$

No· of· atoms on the product side is,

$Zn\ =\ 1\ ,\ \ SO_4\ =\ 1\ ,\ \ Li\ =\ 2\ ,\ \ CO_3\ =\ 1$

This reaction is already balanced one· So, no need to balance again·

$5\ )$ $V_2O_5\ +\ CaS\ --->\ CaO\ +\ V_2S_5$

Here,

No· of· atoms on the reactant side is,

$V\ =\ 2\ ,\ \ O\ =\ 5\ ,\ \ Ca\ =\ 1\ ,\ \ S\ =\ 1$

No· of· atoms on the product side is,

$V\ =\ 2\ ,\ \ O\ =\ 1\ ,\ \ Ca\ =\ 1\ ,\ \ S\ =\ 5$

To balance this equation, add $5$ near the $CaS$ and then add $5$ near the $CaO$ following that·

So, we get the balanced chemical reaction as,

$V_2O_5\ +\ 5\ CaS\ --->\ 5\ CaO\ +\ V_2S_5$