Question

I got everything right except I am having an issue finding the answer to "g aluminum nitrite". The balanced equation is Al(NO2)3 + 3 NH4Cl---> AlCl3 + 6 H20. The limiting reactant is NH4Cl. All atomic weight is rounded to 4 significant figures for this periodic table.

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of each substance is present after 125.0 g of aluminum nitrite and 99.5 g of ammonium chloride react completely? __?__g aluminum nitrite 0 g ammonium chloride 83 g aluminum chloride 52.1 g nitrogen 67.02 g water

Answer 1

1)

balanced equation :

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

165 g         160.5 g                   133.34 g      84 g       108.09 g

125 g          99.5 g

here limiting reagent is NH4Cl.

mass of aluminium nitrite = 125 - ((99.5 x 165) / 160.5)

mass of aluminium nitrite    = 22.7 g

ammonium chloride = 0 g

aluminum chloride = 82.7 g

nitrogen = 52.1 g

water = 67.0 g