A chemist adds 26.5 g of ammonium chloride to 10 g of sodium hydroxide according to the reaction below. Assuming complete reaction, which reactant is left in excess and how much remains?

To answer this question, we need to know the complete chemical reaction and have the balanced chemical equation. However, based on typical reactions between ammonium chloride and sodium hydroxide, let's consider the following reaction:

\[ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{NH}_3(g) \]

This chemical reaction is balanced as written, meaning one mole of ammonium chloride reacts with one mole of sodium hydroxide to produce one mole each of sodium chloride, water, and ammonia gas.

Step 1: First, we need to calculate the molar mass of the reactants to convert grams to moles.

- Molar mass of ammonium chloride (\( \text{NH}_4\text{Cl} \)) = 14.01 (N) + (1.01 * 4) (H) + 35.45 (Cl) = 53.49 g/mol - Molar mass of sodium hydroxide (\( \text{NaOH} \)) = 22.99 (Na) + 15.999 (O) + 1.01 (H) = 39.997 g/mol

Step 2: Now, convert the mass of the reactants to moles to see which is the limiting reactant.

- Moles of \( \text{NH}_4\text{Cl} \) = 26.5 g / 53.49 g/mol = 0.495 moles - Moles of \( \text{NaOH} \) = 10 g / 39.997 g/mol = 0.250 moles

Step 3: The balanced equation tells us that \( \text{NH}_4\text{Cl} \) and \( \text{NaOH} \) react in a 1:1 molar ratio. Compare the moles of the reactants:

- Since we have 0.495 moles of \( \text{NH}_4\text{Cl} \) and only 0.250 moles of \( \text{NaOH} \), the sodium hydroxide is the limiting reactant because there are fewer moles of it compared to \( \text{NH}_4\text{Cl} \).

Therefore, \( \text{NH}_4\text{Cl} \) is in excess.

Step 4: Calculate the excess amount of \( \text{NH}_4\text{Cl} \) that remains after the reaction.

- \( \text{NH}_4\text{Cl} \) remaining = moles of \( \text{NH}_4\text{Cl} \) - moles of \( \text{NaOH} \) used = 0.495 moles - 0.250 moles = 0.245 moles

Step 5: Convert the excess moles of \( \text{NH}_4\text{Cl} \) back to grams.

- Excess \( \text{NH}_4\text{Cl} \) in grams = 0.245 moles * 53.49 g/mol = 13.10 g

Answer: Ammonium chloride is the reactant left in excess, and 13.10 grams of it remains after the reaction.