Question

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of each substance is present after 122.9 g of aluminum nitrite and 97.4 g of ammonium chloride react completely?

Answer 1

Reaction is

Al(NO2)3 + 3 NH4Cl ...............> AlCl3 + 3 N2 + 6 H2O

...1 equiv .........3 equiv

we have

122.9 g of aluminum nitrite and 97.4 g of ammonium chloride.

mole of Al(NO2)3 = 122.9 g / 164.998 g / mole = 0.7449 mole.

and

mole of NH4Cl = 97.4 g / 53.491 g/mole = 1.821 mole.

NH4Cl is the limiting reactant.

thus

after the reaction,

mass of NH4Cl present = 0.00 g (answer)

and

mass of Al(NO2)3 present = 0.7449 - (1.821 / 3) = 0.1379 mole = 0.1379 mole * 164.998 g / mole = 22.8 g (answer)