# Problem: 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?

###### FREE Expert Solution

Balanced equation: Al(NO2)3(aq) + 3 NH4Cl(aq) → AlCl3(aq) + 3 N2(g) + 6 H2O(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 AlCl3 (any of the products can be used). In order to do that, we need to determine the molar mass of Al(NO2)3, NH4Cl, and AlCl3.

The molar mass of AlCl3 is:

AlCl3             1 Al × 26.98 g/mol Al = 26.98 g/mol

3 Cl × 35.45 g/mol Cl = 106.35g/mol

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Sum = 133.33 g/mol

For Al(NO2)3: From the balanced equation, 1 mol of Al(NO2)3 forms 1 mol of AlCl3. The molar mass of Al(NO2)3 is:

Al(NO2)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

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Sum = 165.01 g/ 1mol

The mass of AlClformed by 72.5 g Al(NO2)is:

58.58 g AlCl3

For NH4Cl: From the balanced equation, 3 moles of NH4Cl forms 1 mol of AlCl3. The molar mass of NH4Cl is:

NH4Cl           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

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Sum = 53.492 g/mol

The mass of AlClformed by 58.6 g NH4Cl is:

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###### Problem Details

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?