What mass of aluminum chloride should be taken for the reaction with silver nitrate in order to obtain

What mass of aluminum chloride should be taken for the reaction with silver nitrate in order to obtain a precipitate weighing 58 g, if the product yield is 70%?

First, let’s draw up the reaction equation:
AlCL3 + 3 AgNO3 = 3AgCl + Al (NO3) 3
The insoluble precipitate in our case is the silver chloride salt
Now we will calculate the molar masses of aluminum chloride and silver chloride, according to which we will carry out further calculations:
M (AlCl3) = M (Al) + 3 M (Cl) = 26 + 3 * 35.5 = 132.5 g / mol
M (AgCl) = M (Ag) + M (Cl) = 108 + 35.5 = 143.5 g / mol
We take the values ​​of atomic masses from the Periodic Table of D.I. Mendeleev.
Now we make up the proportion:
m (AlCl3) = (M (AlCl3) * m (AgCl)) / M (AgCl)
To calculate, we need to know the theoretical mass, for which we use the formula to calculate the yield of the reaction product:
n = (m practice / m theory) * 100% = 70%, whence
m theory = m practice / n = 82.9 g
Now let’s calculate the mass of aluminum chloride:
m (AlCl3) = (M (AlCl3) * m (AgCl)) / M (AgCl) = (132.5 * 82.9) / 143.5 = 76.5 g
Answer: the mass of aluminum chloride, which must be taken, is 76.5 g.