Silver nitrate and iron (Ill) chloride are reacted. 27.0 g silver nitrate and 43.5 g iron (III) chloride are used in the reaction. Using the lmiung reactant, calculate how many grams of silver chloride are produced.

Answer 1

First, we need to determine the limiting reactant in the reaction.

1. Calculate the molar mass of each compound:
- Silver nitrate (AgNO3): 107.87 g/mol (Ag) + 14.01 g/mol (N) + (3 x 16.00 g/mol) = 169.87 g/mol
- Iron (III) chloride (FeCl3): 55.85 g/mol (Fe) + (3 x 35.45 g/mol) = 162.20 g/mol

2. Calculate the number of moles of each compound:
- Moles of AgNO3 = 27.0 g / 169.87 g/mol = 0.1589 mol
- Moles of FeCl3 = 43.5 g / 162.20 g/mol = 0.2683 mol

3. Determine the stoichiometry of the reaction:
The balanced chemical equation for the reaction between Silver nitrate and Iron (III) chloride is:
2 AgNO3 + FeCl3 → 2 AgCl + Fe(NO3)3

This equation shows that 2 moles of AgNO3 react with 1 mole of FeCl3 to produce 2 moles of AgCl.

4. Determine the limiting reactant:
Since the stoichiometry is 2:1 for AgNO3:FeCl3, and there are fewer moles of AgNO3 than FeCl3, AgNO3 is the limiting reactant.

5. Calculate the theoretical yield of AgCl:
From the stoichiometry, 0.1589 mol AgNO3 will produce 0.1589 mol AgCl.

6. Convert moles of AgCl to grams:
- Molar mass of AgCl: 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol
- Theoretical yield of AgCl = 0.1589 mol * 143.32 g/mol = 22.8 g

Therefore, if 27.0 g of silver nitrate and 43.5 g of iron (III) chloride are reacted, 22.8 g of silver chloride are produced.