Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of each substance is present after 111.6 g of aluminum nitrite and 86.1 g of ammonium chloride react completely?

The answer to your question is

 Al(NO₂)₃ = 23.08 g; NH₄Cl = 0;   AlCl₃ = 71,62; N₂ = 45.06 g;  H₂O = 57.94 g

Explanation:

Data

Al(NO₂)₃ = 111.6 g

NH₄Cl = 86.1 g

AlCl₃ = ?

N₂ = ?

H₂O = ?

Balanced chemical reaction

                Al(NO₂)₃  +  3NH₄Cl  ⇒   AlCl₃  +  3N₂  + 6H₂O

Process

1.- Calculate the molar mass of the reactants and products

Al(NO₂)₃ = 27 + (14 x 3) + (16 x 6) = 165 g

3NH₄Cl = 3[14 + 35.5 + 4] = 160.5 g

AlCl₃ = 27 + (35.5 x 3) = 133.5 g

3N₂ = 3(14 x 2) = 84 g

6H₂O = 6(2 + 16) = 108 g

2.- Calculate the limiting reactant

theoretical yield = 165/160.5 = 1.03

experimental yield = 111.6/86.1 = 1.3

From this information we conclude that the limiting reactant is NH₄Cl because the experimental yield was higher.

3.- Calculate the mass of the products

             160.5 g of NH₄Cl 133.5 g of AlCl₃

               86.1 g of NH₄Cl    x

               x = (86.1 x 133.5)/160.5

               x = 71.62 g of AlCl₃

            160.5 g of NH₄Cl 84 g of N₂

               86.1 g of NH₄Cl    x

               x = (86.1 x 84)/160.5

               x = 45.06 g of AlCl₃

             160.5 g of NH₄Cl 108 g of H₂O

               86.1 g of NH₄Cl    x

               x = (86.1 x 108)/160.5

               x = 57.94 g of AlCl₃

              160.5 g of NH₄Cl 165 g of Al(NO₂)₃

               86.1 g of NH₄Cl    x

               x = (86.1 x 165)/160.5

               x = 88.51 g of AlCl₃

Mass of Al(NO₂)₃ = 111.6 - 88.51

                            = 23.08 g

There remains 23.0 grams Al(NO2)3

Mass of products:

Mass AlCl3 =  71.6 grams

Mass N2 = 45.1 grams

Mass H2O = 58.0 grams

Explanation:

Step 1: Data given

Mass of aluminium nitrite= 111.6 grams

Molar mass aluminium nitrite = 165.0 g/mol

Mass of ammonium chloride = 86.1 grams

Molar mass of ammonium chloride = 53.49 g/mol

Step 2: The balanced equation

Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O

Step 3: Calculate moles Al(NO2)3

Moles Al(NO2)3 = mass / molar mass

Moles Al(NO2)3 = 111.6 grams / 165.0 g/mol

Moles Al(NO2)3 = 0.6764 moles

Step 4: Calculate moles NH4Cl

Moles NH4Cl = 86.1 grams / 53.49 g/mol

Moles NH4Cl = 1.61 moles

Step 5: Calculate limiting reactant

For 1 mol Al(NO2)3 we need 3 moles NH4Cl to produce 1 mol AlCl3, 3 moles N2 and 6 moles H2O

NH4Cl is the limiting reactant. IT will completely be consumed (1.61 moles).

Al(NO2)3 is in excess. There will react 1.61/3 = 0.537 moles

There will remain 0.6764 - 0.537 = 0.1394 moles

This is 0.1394 * 165.0 g/mol = 23.0 grams

Step 6: calculate moles of products

For 3 mol NH4Cl we need 1 mol Al(NO2)3 to produce 1 mol AlCl3, 3 moles N2 and 6 moles H2O

For 1.61 moles of Al(NO2)3 we'll have:

1.61/ 3 = 0.537 moles AlCl3

1*1.61 = 1.61 moles N2

2*1.61 = 3.22 moles H2O

Step 7: Calculate mass of products

Mass =moles * molar mass

Mass AlCl3 = 0.537 moles *133.34 g/mol = 71.6 grams

Mass N2 = 1.61 moles * 28.0 g/mol = 45.1 grams

Mass H2O = 9.66 moles * 18.02 g/mol = 58.0 grams

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given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g