The following reaction occurs in an aqueous acid solution:

no3- + I- -> IO3- + NO2
In the balanced equation, the coefficient of water is
A. 4
B. 3
C. 1
D. 5
E. 2

To determine the coefficient of water in the given equation, we need to balance the equation first.

Step 1: Assign coefficients to each molecule or ion in the equation:

no3- + I- -> IO3- + NO2

Step 2: Start with the most complex molecule/ion and balance its elements first. In this case, let's start with IO3-. It has one iodine atom, so we put a coefficient of 1 in front of IO3-:

no3- + I- -> 1 IO3- + NO2

Step 3: Next, balance the iodine atoms. There is one iodine atom on the reactant side (I-) and one on the product side (IO3-). Therefore, iodine is already balanced.

no3- + 1 I- -> 1 IO3- + NO2

Step 4: Now, balance the nitrogen atoms. There is one nitrogen atom on both sides of the equation, so nitrogen is already balanced.

no3- + 1 I- -> 1 IO3- + NO2

Step 5: Finally, balance the oxygen atoms. In this reaction, there are three oxygen atoms on the reactant side (one in NO3- and two in I-) and four oxygen atoms on the product side (three in IO3- and one in NO2). To balance the oxygen atoms, we need to add a coefficient of 2 in front of NO3- and 2 in front of I-:

2 no3- + 2 I- -> 1 IO3- + NO2

Now, let's analyze the balanced equation. We see that for each NO3- ion, we need two water molecules (H2O) to balance the oxygen atoms on the reactant side. Therefore, the coefficient of water (H2O) is twice the coefficient of NO3-:

Coefficient of water = 2 * Coefficient of NO3-

In this balanced equation, the coefficient of NO3- is 2. Therefore, the coefficient of water is 2 * 2 = 4.

So, the correct answer is A. 4.