Respuesta :
Answer:
Absolutely, let's calculate the theoretical yield of ZrF4 produced in the reaction:
**1. Balanced Chemical Equation:**
The balanced chemical equation for the reaction is:
Zr(SO4)2 (aq) + 4AgF (s) -> 2Ag2SO4 (aq) + ZrF4 (s)
This equation tells us that 4 moles of AgF react to produce 1 mole of ZrF4.
**2. Moles of AgF:**
We are given the number of particles of AgF: 6.11 * 10^23 particles. To convert this to moles, we need Avogadro's constant:
* Avogadro's constant (N_A) = 6.02214076 x 10^23 particles/mol
Moles of AgF = (Number of particles) / (Avogadro's constant)
Moles of AgF = (6.11 * 10^23 particles) / (6.02214076 x 10^23 particles/mol)
Moles of AgF ≈ 1.016 mol
**3. Moles of ZrF4 (Theoretical Yield):**
From the balanced equation, we know that 4 moles of AgF react to produce 1 mole of ZrF4. Since we have 1.016 moles of AgF (assuming the reaction goes to completion), we can determine the theoretical yield of ZrF4 using the following proportion:
Moles of ZrF4 / Moles of AgF = 1 / 4
Moles of ZrF4 = (Moles of AgF) * (1 / 4)
Moles of ZrF4 = (1.016 mol) * (1 / 4)
Moles of ZrF4 ≈ 0.254 mol
**4. Grams of ZrF4:**
Now that we know the theoretical yield in moles, we can convert it to grams using the molar mass of ZrF4:
* Molar mass of ZrF4 = 209.22 g/mol (grams per mole of ZrF4)
Grams of ZrF4 = (Moles of ZrF4) * (Molar mass of ZrF4)
Grams of ZrF4 = (0.254 mol) * (209.22 g/mol)
Grams of ZrF4 ≈ 53.07 grams
Therefore, the theoretical yield of ZrF4 produced in the reaction is approximately 53.07 grams.
