In the following equation , 417.96 g of Bi reacts completely with 200 g of F . How many grams of BIF (to the third power) are formed?
molar mass of Bi is 208.98 g
... so there are 2 moles of Bi
6 moles of F will react to form BiF₃
the molar mass of F is 19.0 g
add the mass of the Bi and the mass of the F to find the grams of BiF₃
To find the amount of BIF₃ formed, we need to calculate the limiting reactant first. The method involves comparing the stoichiometry of the reactants and determining which one will react completely, limiting the reaction.
The balanced chemical equation for the reaction is:
2Bi + 3F₂ → 2BIF₃
Now let's calculate the moles of each reactant.
For Bi:
Molar mass of Bi = 208.98 g/mol
Moles of Bi = mass of Bi / molar mass of Bi
= 417.96 g / 208.98 g/mol
≈ 2.00 mol
For F:
Molar mass of F₂ = 19.00 g/mol
Moles of F = mass of F / molar mass of F₂
= 200 g / 19.00 g/mol
≈ 10.53 mol
Using the stoichiometry of the balanced equation, we can determine the moles ratio of Bi to F:
2 moles of Bi: 3 moles of F₂
To find the limiting reactant, we compare the moles of Bi and F using their stoichiometry ratio:
2.00 mol Bi * (3 mol F₂ / 2 mol Bi) = 3.00 mol F (theoretical)
Since the moles of F (10.53 mol) are greater than the theoretical moles (3.00 mol), Bi is the limiting reactant.
Now, using the stoichiometry, let's find the moles of BIF₃ formed:
2 moles of Bi = 2 moles of BIF₃
Moles of BIF₃ = 2.00 mol Bi × (2 mol BIF₃ / 2 mol Bi)
= 2.00 mol BIF₃
Finally, let's calculate the mass of BIF₃ formed:
Molar mass of BIF₃ = (1 mol Bi × molar mass of Bi) + (3 mol F₂ × molar mass of F)
= (1 mol × 208.98 g/mol) + (3 mol × 19.00 g/mol)
= 208.98 g + 57.00 g
= 265.98 g/mol
Mass of BIF₃ = moles of BIF₃ × molar mass of BIF₃
= 2.00 mol × 265.98 g/mol
≈ 532 g
Therefore, approximately 532 grams of BIF₃ are formed.
To find the number of grams of BIF (to the third power) formed in the reaction, we need to determine the balanced chemical equation and use stoichiometry to calculate the mass.
1. Start by writing the balanced chemical equation for the reaction between Bi and F.
Bi + 3F₂ → BIF₃
2. Calculate the molar mass of Bi, F₂, and BIF₃.
- Molar mass of Bi = 208.98 g/mol
- Molar mass of F₂ = 38.00 g/mol
- Molar mass of BIF₃ = (1 × molar mass of Bi) + (3 × molar mass of F) = (1 × 208.98 g/mol) + (3 × 38.00 g/mol) = 367.94 g/mol
3. Convert the given masses of Bi and F to moles.
- Moles of Bi = Mass of Bi / Molar mass of Bi = 417.96 g / 208.98 g/mol
- Moles of F = Mass of F / Molar mass of F₂ = 200 g / 38.00 g/mol
4. Determine the limiting reactant by comparing the moles of Bi and F. The reactant that produces fewer moles of the product (BIF₃) will be the limiting reactant.
- Limiting reactant = Reactant with fewer moles
5. Use the stoichiometry of the balanced equation to determine how many moles of BIF₃ will be formed.
- From the balanced equation, 1 mole of Bi reacts with 3 moles of F₂ to form 1 mole of BIF₃.
6. Convert moles of BIF₃ to grams using its molar mass.
- Mass of BIF₃ = Moles of BIF₃ × Molar mass of BIF₃
By following these steps, you can determine the grams of BIF₃ formed in the reaction. Make sure to substitute the values for the molar masses and calculate the moles accurately to get the correct answer.