Question

Mn2+(aq) + nabio3(s) → bi3+(aq) + mno4−(aq) + na+(aq) how many water molecules are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)

Answer 1

Answer : The number of water molecules are, 7

Explanation :

The given unbalanced chemical reaction is,

$Mn^{2+}+NaBiO_3\rightarrow Bi^{3+}+MnO_4^-+Na^+$

First we have to separate into half reaction. The two half reactions are:

$Mn^{2+}\rightarrow MnO_4^-$

$NaBiO_3\rightarrow Na^++Bi^{3+}$

Now we have to balance the half reactions, we get

$Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^++5e^-$     ............(1)

$NaBiO_3+6H^++2e^-\rightarrow Na^++Bi^{3+}+3H_2O$      ........(2)

Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2 and equation (2) by 5, we get

$2Mn^{2+}+8H_2O\rightarrow 2MnO_4^-+16H^++10e^-$

$5NaBiO_3+30H^++10e^-\rightarrow 5Na^++5Bi^{3+}+15H_2O$

Now we have to add both the half reactions, we get the final balanced chemical reaction.

$2Mn^{2+}+5NaBiO_3+14H^+\rightarrow 5Bi^{3+}+2MnO_4^-+5Na^++7H_2O$

From the final balanced chemical reaction, we conclude that there are 7 number of water molecules present in this reaction on product side.

Hence, the number of water molecules are, 7

Answer 2

Mn²⁺ + 4H2O -----> MnO4⁻ + 8H⁺ +5e⁻                  /*2
NaBiO3 +6H⁺ +2e⁻ -----> Bi³⁺ + Na⁺ + 3H2O        /*5
2Mn²⁺ + 5 NaBiO3+8H2O+30H⁺ --->  2MnO4⁻ +5Bi³⁺ + 5Na⁺ +16H⁺ +15H2O

2Mn²⁺ + 5 NaBiO3+14H⁺ --->  2MnO4⁻ +5Bi³⁺ + 5Na⁺  +7H2O

There are 7 water molecules in this reaction.