Consider the reaction of the nitrate ion with the tin(II) ion in aqueous solution to produce nitrogen monoxide gas and the tin(IV) ion. When the chemical equation for this reaction is balanced in acidic media correctly with the smallest whole-number coefficients, what is the coefficient of water

2NO3-(aq) + 3Sn2+(aq) + 8H+(aq) —> 2NO(g) + 3Sn4+(aq) + 4H2O(l)

The coefficient of water (H2O) is 4

Explanation:

Step 1:

The unbalanced equation:

NO3-(aq) + Sn2+(aq) —> NO + Sn4+

Step 2:

The equation in an acid medium. This is illustrated below:

H+ indicating the acid will be added and H2O will be produced on the other side of the equation as shown below:

NO3-(aq) + Sn2+(aq) + H+(aq) —> NO(g) + Sn4+(aq) + H2O(l)

Step 3:

Balancing the equation.

NO3-(aq) + Sn2+(aq) + H+(aq) —> NO(g) + Sn4+(aq) + H2O(l)

The above equation can be balance as follow:

There are 3 atoms of O on the left side of the equation and a total of 2 atoms on the right side. It can be balance by putting 2 in front of NO3-, 2 in front of NO and 4 in front of H2O as shown below:

2NO3-(aq) + Sn2+(aq) + H+(aq) —> 2NO(g) + Sn4+(aq) + 4H2O(l)

Now we have 8 atoms of H on the right side and 1 atom on the left side. It can be balance by putting 8 in front of H+ as shown below:

2NO3-(aq) + Sn2+(aq) + 8H+(aq) —> 2NO(g) + Sn4+(aq) + 4H2O(l)

Considering the charge, we have a total charge of +8 on the left side and +4 on the right side. It can be balance by putting 3 in front of Sn2+ and 3 in front of Sn3+ as shown below:

2NO3-(aq) + 3Sn2+(aq) + 8H+(aq) —> 2NO(g) + 3Sn4+(aq) + 4H2O(l)

Now the equation is balanced.

The coefficient of water (H2O) in the equation above is 4

the ratio of the conjugate base to the acid required to attain the desired ph.