116.7 g of Aluminum nitrite and 193.2 g of ammonium chloride react to form aluminum chloride, nitrogen, and water.  How many grams of the excess reagent remain after the reaction? Enter to 2 decimal places.

To determine how many grams of the excess reagent remain after the reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.

The reaction between aluminum nitrite \((\text{Al(NO}_2)_3)\) and ammonium chloride \((\text{NH}_4\text{Cl})\) can be written as follows:

\[
\text{Al(NO}_2)_3 + 3 \text{NH}_4\text{Cl} \rightarrow \text{AlCl}_3 + 3 \text{N}_2 + 6 \text{H}_2\text{O}
\]

### Step 2: Calculate the molar masses of the reactants.

- **Aluminum nitrite (Al(NO\(_2\))\(_3\))**:
  \[
  \text{Al} = 26.98 \, \text{g/mol}, \, \text{N} = 14.01 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}
  \]
  \[
  \text{Molar mass of Al(NO}_2)_3 = 26.98 + 3(14.01 + 2 \times 16.00) = 26.98 + 3(46.01) = 165.01 \, \text{g/mol}
  \]

- **Ammonium chloride (NH\(_4\)Cl)**:
  \[
  \text{N} = 14.01 \, \text{g/mol}, \, \text{H} = 1.008 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol}
  \]
  \[
  \text{Molar mass of NH}_4\text{Cl} = 14.01 + 4(1.008) + 35.45 = 53.49 \, \text{g/mol}
  \]

### Step 3: Convert the masses of reactants to moles.

- Moles of aluminum nitrite:
  \[
  \frac{116.7 \, \text{g}}{165.01 \, \text{g/mol}} = 0.707 \, \text{mol}
  \]

- Moles of ammonium chloride:
  \[
  \frac{193.2 \, \text{g}}{53.49 \, \text{g/mol}} = 3.614 \, \text{mol}
  \]

### Step 4: Determine the limiting reagent.

From the balanced equation, 1 mole of Al(NO\(_2\))\(_3\) reacts with 3 moles of NH\(_4\)Cl. To react with 0.707 moles of Al(NO\(_2\))\(_3\), the required moles of NH\(_4\)Cl are:
\[
0.707 \, \text{mol Al(NO}_2)_3 \times 3 = 2.121 \, \text{mol NH}_4\text{Cl}
\]

Since we have 3.614 moles of NH\(_4\)Cl, which is more than 2.121 moles, ammonium chloride is in excess, and aluminum nitrite is the limiting reagent.

### Step 5: Calculate the amount of excess ammonium chloride remaining.

The amount of ammonium chloride reacted is 2.121 moles, so the remaining moles are:
\[
3.614 - 2.121 = 1.493 \, \text{mol NH}_4\text{Cl}
\]

### Step 6: Convert the remaining moles of ammonium chloride to grams.

\[
1.493 \, \text{mol} \times 53.49 \, \text{g/mol} = 79.88 \, \text{g}
\]

### Final Answer:
There are **79.88 g** of ammonium chloride remaining after the reaction.

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you determine the limiting reagent in a chemical reaction?
2. What are the products formed in this reaction, and how are they derived from the reactants?
3. How do you calculate the theoretical yield of a product in a chemical reaction?
4. What is the significance of the stoichiometric ratio in a balanced chemical equation?
5. How would you calculate the percent yield of a reaction?

**Tip:** Always ensure the chemical equation is balanced before starting calculations for stoichiometry.

116.7 g of Aluminum nitrite and 193.2 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of the excess reagent remain after the reaction? Enter to 2 decimal places.

Learn how to calculate the excess reagent in a chemical reaction between 116.7 g of aluminum nitrite and 193.2 g of ammonium chloride. This problem covers stoichiometry, molar mass calculation, and limiting reagent concepts. Suitable for undergraduate chemistry students.

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