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. How marry grams or the excess resctant will remain after the resction is comolete?
To find out which reactant is limiting and which is in excess, we need to first determine the limiting reactant by calculating the moles of each reactant.
1. Calculate the moles of each reactant:
- Moles of silver nitrate (AgNO3):
27.0 g / 169.87 g/mol = 0.159 mol
- Moles of iron (III) chloride (FeCl3):
43.5 g / 162.20 g/mol = 0.268 mol
2. Write the balanced chemical equation for the reaction:
2 AgNO3 + FeCl3 → 2 AgCl + Fe(NO3)3
3. Determine the stoichiometry of the reaction based on the balanced chemical equation:
For every 2 moles of AgNO3, 1 mole of FeCl3 is required.
4. Calculate the maximum amount of FeCl3 that can react with 0.159 mol of AgNO3:
0.159 mol AgNO3 x (1 mol FeCl3 / 2 mol AgNO3) = 0.0795 mol FeCl3
5. Compare the actual amount of FeCl3 used (0.268 mol) with the maximum amount that can react (0.0795 mol):
Since 0.268 mol of FeCl3 is greater than 0.0795 mol, FeCl3 is in excess.
6. Calculate the amount of excess FeCl3 remaining after the reaction:
0.268 mol - 0.0795 mol = 0.1885 mol
7. Convert the excess mol of FeCl3 to grams:
0.1885 mol FeCl3 x 162.20 g/mol = 30.56 g
Therefore, 30.56 grams of excess FeCl3 will remain after the reaction is complete.