For the same reaction above, if 32 grams of AlCl3 is combined with excess AgNO3, how many grams of AgCl can be made

103.2 g of AgCl can be made in the reaction

Explanation:

We state the reaction to analyse the problem:

AlCl₃ + 3AgNO₃ →  3AgCl  +  Al(NO₃)₃

We convert the mass of chloride to moles:

32 g . 1mol/ 133.34g = 0.240 moles

Ratio is 1:3. See the stoichiometry

1 mol of aluminum chloride can produce 3 moles of AgCl

Then, 0.240 will produce (0.240 . 3)/1= 0.720 moles.

We convert the moles to mass: 0.720 mol . 143.32g/1mol = 103.2 g