3) If 3.26 moles of sodium hydroxide react with 9.87 moles of potassium dichromate.Determine the moles of potassium hydroxide and identify the limiting and excess reactant. Balanced Equation: __

## Step 1<br />The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and potassium dichromate (K2Cr2O7) is:<br /><br />### \(7NaOH + K2Cr2O7 \rightarrow 2Na2CrO4 + 2KOH + 4H2O\)<br /><br />This equation tells us that 7 moles of sodium hydroxide react with 1 mole of potassium dichromate to produce 2 moles of potassium hydroxide.<br /><br />## Step 2<br />We are given that 3.26 moles of sodium hydroxide and 9.87 moles of potassium dichromate are present. We need to determine which reactant is the limiting reactant, which is the reactant that will be completely consumed in the reaction.<br /><br />## Step 3<br />To find the limiting reactant, we compare the ratio of the moles of sodium hydroxide to potassium dichromate with the ratio in the balanced equation. <br /><br />## Step 4<br />The ratio of sodium hydroxide to potassium dichromate in the balanced equation is 7:1. However, the ratio of sodium hydroxide to potassium dichromate in the given problem is 3.26:9.87, which is less than the ratio in the balanced equation. This indicates that sodium hydroxide is the limiting reactant.<br /><br />## Step 5<br />Since sodium hydroxide is the limiting reactant, potassium dichromate is the excess reactant.<br /><br />## Step 6<br />To find the moles of potassium hydroxide produced, we use the stoichiometry of the reaction. According to the balanced equation, 7 moles of sodium hydroxide produce 2 moles of potassium hydroxide. Therefore, 3.26 moles of sodium hydroxide will produce \( \frac{2}{7} \times 3.26 = 0.93 \) moles of potassium hydroxide.