Calculate:
In the following equation, 417.96 g of Bi (bismuth) reacts completely with 200 g of F (fluorine.) How many grams of BiF (little three below F) are formed?
2 Bi + 3 F (little 2 under F) ---> 2 BiF (little three below F)
This is a limiting reagent (LR) problem and I know that because amounts are given for BOTH reactants. I solve these the long way--there is a shorter way but I think it's harder to explain, especially by typing.
2Bi + 3F2 ==> 2BiF3
mols Bi = grams/molar mass = ?
mols F2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Bi to mols BiF3.
Do the same to convert mols F2 to mols BiF3.
It is quite likely that the two values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that smaller value is the LR.
Take the smaller value and convert that to grams. g = mols x molar mass. = ?grams BiF3
500
To determine the number of grams of BiF (bismuth fluoride) formed, we can use the concept of stoichiometry.
Step 1: Start with the given mass of Bi (bismuth) and F (fluorine).
Bi = 417.96 g
F = 200 g
Step 2: Calculate the molar mass of the reactants and product.
The molar mass of Bi is 208.98 g/mol, and the molar mass of F is 18.998 g/mol.
Step 3: Convert the given mass of Bi and F to moles.
moles of Bi = mass/molar mass = 417.96 g / 208.98 g/mol = 2 mol
moles of F = mass/molar mass = 200 g / 18.998 g/mol = 10.53 mol (rounded to two decimal places)
Step 4: Determine the molar ratio between Bi and BiF in the balanced equation.
From the balanced equation: 2 Bi : 2 BiF
The molar ratio is 2:2, or 1:1.
Step 5: Calculate the moles of BiF formed.
Since the molar ratio is 1:1, the moles of BiF formed will be equal to the moles of Bi used (2 mol).
Step 6: Convert the moles of BiF to grams.
mass of BiF = moles of BiF x molar mass of BiF
The molar mass of BiF is (208.98 g/mol + 18.998 g/mol) = 227.978 g/mol.
mass of BiF = 2 mol x 227.978 g/mol = 455.956 g (rounded to three decimal places)
Therefore, 455.956 grams of BiF are formed when 417.96 g of Bi reacts completely with 200 g of F.
To calculate the grams of BiF3 formed, we need to use stoichiometry, which is the mathematical relationship between the reactants and products in a chemical equation.
First, we need to determine the molar mass of each element involved in the reaction. The molar masses are as follows:
- Bi (bismuth) molar mass = 208.98 g/mol
- F (fluorine) molar mass = 18.998 g/mol
- BiF3 (bismuth trifluoride) molar mass = 263.98 g/mol
Next, we need to determine the number of moles of Bi and F in the reaction. To do this, we divide the given masses of Bi and F by their respective molar masses:
- Moles of Bi = 417.96 g / 208.98 g/mol = 1.998 moles
- Moles of F = 200 g / 18.998 g/mol = 10.530 moles
We can see from the balanced chemical equation that the ratio between Bi and BiF3 is 2:2 (1:1) and the ratio between F and BiF3 is 3:2. Therefore, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the moles of Bi and F. The mole ratio between Bi and F is 2:3. For every 2 moles of Bi, we need 3 moles of F. Therefore, the limiting reactant is F, as we have excess Bi.
Now, we can calculate the moles of BiF3 that can be formed using the limiting reactant, which is F. The mole ratio between F and BiF3 is 3:2:
- Moles of BiF3 = (10.530 moles of F) * (2 moles of BiF3 / 3 moles of F) = 7.02 moles
Finally, we can determine the mass of BiF3 formed by multiplying the moles of BiF3 by its molar mass:
- Mass of BiF3 = 7.02 moles * 263.98 g/mol = 1851.7596 g
Therefore, approximately 1851.76 grams of BiF3 are formed in the reaction.