🤓 Based on our data, we think this question is relevant for Professor Stec's class at UIC.

**Balanced equation: Al(NO _{2})_{3(aq)} + 3 NH_{4}Cl_{(aq)} → AlCl_{3(aq)} + 3 N_{2(g)} + 6 H_{2}O_{(l)}**

We shall determine which is the limiting and excess reactant.

Notice that we are given the mass of both reactants: this means we need to determine the *limiting reactant*, which is the reactant that forms the less amount of product. This is because once the limiting reactant is all used up, the reaction can no longer proceed and make more products.

This means the limiting reactant determines the maximum mass of the product formed.

We need to perform a *mole-to-mole comparison between* each reactant and AlCl_{3} (any of the products can be used). In order to do that, we need to determine the molar mass of Al(NO_{2})_{3}, NH_{4}Cl, and AlCl_{3}.

The molar mass of AlCl_{3} is:

**AlCl**_{3} 1 Al × 26.98 g/mol Al = 26.98 g/mol

** 3 Cl × 35.45 g/mol Cl = 106.35g/mol**

** ^{_______________________________________________}**

*Sum = **133.33 g/mol*

For Al(NO_{2})_{3}: From the balanced equation, 1 mol of Al(NO_{2})_{3} forms 1 mol of AlCl_{3}. The molar mass of Al(NO_{2})_{3} is:

**Al(NO**_{2}**)**_{3}_{ }**1**** Al x ****26.98 ****g/mol Al = 26.98 g/mol **

**3**** N x ****14.01**** g/mol N = 42.03 g/mol**

**6**** O x ****16.00**** g/mol O = 96.00 g/mol**

^{____________________________________________}

** Sum = **

The mass of AlCl_{3 }formed by 72.5 g Al(NO_{2})_{3 }is:

$\mathbf{72}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Al}{\left({\mathbf{NO}}_{\mathbf{2}}\right)}_{\mathbf{3}}}\mathbf{\times}\frac{\mathbf{1}{\mathbf{}}\overline{)\mathbf{mol}\mathbf{}\mathbf{Al}{\left({\mathbf{NO}}_{\mathbf{2}}\right)}_{\mathbf{3}}}}{\mathbf{165}\mathbf{.}\mathbf{01}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Al}{\left({\mathbf{NO}}_{\mathbf{2}}\right)}_{\mathbf{3}}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{AlCl}}_{\mathbf{3}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Al}{\left({\mathbf{NO}}_{\mathbf{2}}\right)}_{\mathbf{3}}}}\mathbf{\times}\frac{\mathbf{133}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{AlCl}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{AlCl}}_{\mathbf{3}}}}\mathbf{=}$**58.58 g AlCl**_{3}

For NH_{4}Cl: From the balanced equation, 3 moles of NH_{4}Cl forms 1 mol of AlCl_{3}. The molar mass of NH_{4}Cl is:

**NH**_{4}**Cl ****1**** N x ****14.01**** g/mol N = 14.01 g/mol**

** 4**** H x ****1.008**** g/mol H = 4.032 g/mol**

** 1**** Cl x ****35.45**** g/mol Cl = 35.45 g/mol**

** ^{_____________________________________________}**

*Sum = **53.492 g/mol*

The mass of AlCl_{3 }formed by 58.6 g NH_{4}Cl is:

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance are present after 72.5 g of aluminum nitrite and 58.6 g of ammonium chloride react completely?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Stec's class at UIC.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.