if 22.0 aluminum chloride is added to 26.0 g of silver nitrate, how many g of that solid product is expected to form

21.95 g of AgCl.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

AlCl3(aq) + 3AgNO3(aq) —> 3AgCl(s) + Al(NO3)3(aq)

Next, we shall determine the masses of AlCl3 and AgNO3 that reacted and the mass of AgCl produced from the balanced equation. This is illustrated below:

Molar mass of AlCl3 = 27 + (3×35.5) = 133.5 g/mol

Mass of AlCl3 from the balanced equation = 1 × 133.5 = 133.5 g

Molar mass of AgNO3 = 108 + 14 + (3×16) = 170 g/mol

Mass of AgNO3 from the balanced equation = 3 × 170 = 510 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 3 × 143.5 = 430.5 g

From the balanced equation above,

133.5 g of AlCl3 reacted with 510 g of AgNO3 to produce 430.5 g of AgCl.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

133.5 g of AlCl3 reacted with 510 g of AgNO3.

Therefore,

22 g of AlCl3 will react with =

(22 × 510)/133.5 = 84.04 g of AgNO3.

Thus, we can see that it will require a higher mass (i.e 84.04 g) of AgNO3 than what was given (i.e 26 g) to react completely with 22 g of AlCl3. Therefore, AgNO3 is the limiting reactant and AlCl3 is the excess reactant.

Finally, we shall determine the mass of the solid product (AgCl) produced from the reaction.

In this case the limiting reactant will be used because, it will produce the maximum yield of the reaction since all of it is used up in the reaction.

The limiting reactant is AgNO3 and the mass of AgCl produced can be obtained as follow:

From the balanced equation above,

510 g of AgNO3 reacted to produce 430.5 g of AgCl.

Therefore, 26 g of AgNO3 will react to produce = (26 × 430.5)/510 = 21.95 g of AgCl.

Thus, the mass of the solid product (AgCl) produced is 21.95 g