Question

3 AgNO3 + FeCl3 →3 AgCl + Fe(NO3)3

If you combine 6.60 grams of FeCl3 with an excess of AgNO3, how much AgCl will you form?

Answer 1

To determine the amount of AgCl formed when 6.60 grams of FeCl3 reacts with an excess of AgNO3, we need to use stoichiometry and molar ratios.

First, we need to determine the molar mass of FeCl3 and AgCl:
Molar mass of FeCl3 = atomic mass of Fe + (3 * atomic mass of Cl)
Molar mass of FeCl3 = 55.845 g/mol + (3 * 35.453 g/mol)
Molar mass of FeCl3 = 162.204 g/mol

From the balanced chemical equation, we can see that the molar ratio between FeCl3 and AgCl is 1:3. This means that for every 1 mole of FeCl3, 3 moles of AgCl are produced.

Next, we calculate the number of moles of FeCl3:
Moles of FeCl3 = mass of FeCl3 / molar mass of FeCl3
Moles of FeCl3 = 6.60 g / 162.204 g/mol

Now, using the molar ratio, we can determine the moles of AgCl formed:
Moles of AgCl = Moles of FeCl3 * (3 moles AgCl / 1 mole FeCl3)

Finally, we can calculate the mass of AgCl:
Mass of AgCl = Moles of AgCl * molar mass of AgCl

By following these calculations, you can determine the amount of AgCl formed when 6.60 grams of FeCl3 reacts with an excess of AgNO3.

Answer 2

To determine the amount of AgCl formed, we need to follow the stoichiometry of the balanced equation and calculate the molar amounts of the reactants and products.

First, let's calculate the number of moles of FeCl3 used:

Molar mass of FeCl3 = atomic mass of Fe + (3 * atomic mass of Cl)

= (55.845 g/mol) + (3 * 35.453 g/mol)

= 162.204 g/mol

Moles of FeCl3 = mass of FeCl3 / molar mass of FeCl3

= 6.60 g / 162.204 g/mol

= 0.0407 mol

According to the balanced equation, the ratio of FeCl3 to AgCl is 1:3. Therefore, 1 mol of FeCl3 reacts to form 3 mol of AgCl.

Moles of AgCl formed = 3 * moles of FeCl3

= 3 * 0.0407 mol

= 0.1221 mol

Finally, let's calculate the mass of AgCl formed:

Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl

= 107.868 g/mol + 35.453 g/mol

= 143.321 g/mol

Mass of AgCl formed = moles of AgCl formed * molar mass of AgCl

= 0.1221 mol * 143.321 g/mol

= 17.49 g

Therefore, if you combine 6.60 grams of FeCl3 with an excess of AgNO3, you will form approximately 17.49 grams of AgCl.