Chlorine dioxide `(ClO_(2))` is used now days for water treatment rather than `Cl_(2)` `ClO_(2)` is obtained by passing `(Cl_(2)(g))` into a concentrated solution of sodium chlorite `(NaClO_(2))`. The reaction gives `90%` yield. How many moles of `ClO_(2)` is produced in 3.78 L of 2.0 M `NaClO_(2)(aq)`?

Explanation

To solve the problem of how many moles of chlorine dioxide `(ClO2)` are produced from a given amount of sodium chlorite `(NaClO2)` with a 90% yield, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sodium chlorite and chlorine gas is as follows:
\[ 2 \text{NaClO}_2 + \text{Cl}_2 \rightarrow 2 \text{NaCl} + 2 \text{ClO}_2 \]

### Step 2: Calculate the moles of `NaClO2`
To find the number of moles of sodium chlorite in the solution, we use the formula:
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