Respuesta :
Explanation:
Certainly! Let's calculate the moles of each reactant and determine the limiting and excess reactants.
1. Moles of Cl2:
- Molar mass of Cl: approximately 35.45 g/mol
- Moles of Cl2 = 0.855 g / (2 * 35.45 g/mol) ≈ 0.0121 mol
2. Moles of KBr:
- Molar mass of K: approximately 39.10 g/mol
- Molar mass of Br: approximately 79.90 g/mol
- Moles of KBr = 3.205 g / (39.10 g/mol + 79.90 g/mol) ≈ 0.0258 mol
Now, we compare the moles of Cl2 and KBr:
- Cl2: 0.0121 mol
- KBr: 0.0258 mol
Since Cl2 has fewer moles than KBr, it is the limiting reactant.
Next, let's calculate the moles of products based on the limiting reactant (Cl2):
The balanced chemical equation for the reaction between Cl2 and KBr is needed to determine the stoichiometric ratios.
Assuming the balanced equation is: `Cl2 + 2KBr → 2KCl + Br2`
Now, using stoichiometry:
- Moles of KCl formed = 2 * moles of Cl2 reacted
- Moles of Br2 formed = 1 * moles of Cl2 reacted
Finally, calculate the moles of excess reactant (KBr):
- Moles of excess KBr = Initial moles of KBr - moles used in the reaction (based on stoichiometry)
If you provide the balanced equation, I can help with the final stoichiometric calculations.
