What is my theoretical yield of iron II chloride if I start with 34 grams of iron II bromide?
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To determine the theoretical yield of iron II chloride (FeCl2) from 34 grams of iron II bromide (FeBr2), we need to consider the molar masses of the compounds and their stoichiometry.
1. Determine the molar mass of iron II bromide (FeBr2):
FeBr2 = (atomic mass of Fe) + 2 x (atomic mass of Br)
FeBr2 = (55.845 g/mol) + 2 x (79.904 g/mol)
FeBr2 = 55.845 g/mol + 159.808 g/mol
FeBr2 = 215.653 g/mol
2. Determine the molar mass of iron II chloride (FeCl2):
FeCl2 = (atomic mass of Fe) + 2 x (atomic mass of Cl)
FeCl2 = (55.845 g/mol) + 2 x (35.453 g/mol)
FeCl2 = 55.845 g/mol + 70.906 g/mol
FeCl2 = 126.751 g/mol
3. Calculate the molar ratio between FeBr2 and FeCl2 by comparing their coefficients in the balanced chemical equation:
FeBr2 + Cl2 → FeCl2 + Br2
The coefficient of FeBr2 is 1, and the coefficient of FeCl2 is also 1.
4. Convert the mass of FeBr2 to moles:
Moles of FeBr2 = Mass of FeBr2 / Molar mass of FeBr2
Moles of FeBr2 = 34 g / 215.653 g/mol
Moles of FeBr2 ≈ 0.1579 mol
5. Use the mole ratio to find the moles of FeCl2 produced:
Moles of FeCl2 = Moles of FeBr2 (based on balanced equation) x Stoichiometric ratio
Moles of FeCl2 = 0.1579 mol FeBr2 x (1 mol FeCl2 / 1 mol FeBr2)
Moles of FeCl2 = 0.1579 mol
6. Convert the moles of FeCl2 to grams:
Mass of FeCl2 = Moles of FeCl2 x Molar mass of FeCl2
Mass of FeCl2 = 0.1579 mol x 126.751 g/mol
Mass of FeCl2 ≈ 20 g
Therefore, the theoretical yield of iron II chloride (FeCl2) when starting with 34 grams of iron II bromide (FeBr2) is approximately 20 grams.
To calculate the theoretical yield of iron II chloride, we need to use the stoichiometry of the reaction. The balanced chemical equation for the reaction between iron II bromide (FeBr2) and chlorine gas (Cl2) to form iron II chloride (FeCl2) is:
FeBr2 + Cl2 -> 2FeCl2 + Br2
From the balanced equation, we see that one mole of FeBr2 reacts with one mole of Cl2 to produce two moles of FeCl2.
First, we need to convert the given mass of FeBr2 (34 grams) into moles. To do this, we need to know the molar mass of FeBr2:
Fe: 55.85 g/mol
Br: 79.90 g/mol
Molar mass of FeBr2 = (1 x 55.85 g/mol) + (2 x 79.90 g/mol) = 215.65 g/mol
Number of moles of FeBr2 = mass / molar mass = 34 g / 215.65 g/mol ≈ 0.1579 mol
Now that we know the number of moles of FeBr2, we can determine the theoretical yield of FeCl2. Since two moles of FeCl2 are produced for every one mole of FeBr2, we can multiply the number of moles of FeBr2 by the coefficient ratio:
Theoretical yield of FeCl2 = moles of FeBr2 x (2 moles of FeCl2 / 1 mole of FeBr2)
= 0.1579 mol x 2
= 0.3158 mol
Finally, we can convert the moles of FeCl2 into grams by multiplying by the molar mass of FeCl2:
Molar mass of FeCl2 = (1 x 55.85 g/mol) + (2 x 35.45 g/mol) = 126.75 g/mol
Theoretical yield of FeCl2 = moles of FeCl2 x molar mass of FeCl2
= 0.3158 mol x 126.75 g/mol
≈ 39.97 grams
Therefore, the theoretical yield of iron II chloride is approximately 39.97 grams if you start with 34 grams of iron II bromide.
moles iron(II) bromide = moles/molar mass.
There is one mole iron(II) chloride in one mole iron(II) bromide.
Convert moles iron(II) chloride to grams. g = moles x molar mass. Note: you REALLY should place parentheses around the Roman numerals.