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.
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.