P4 + 6 Br2 -> 4PBr3
In a particular reaction, 20 grams of phosphorus P4, is combined with 50mL liquid bromine ( Density= 2.93 g/ml)
1) Calculate the mass of product in grams for each of the reactants
2) Limiting reagant and Mass Maximum mass of PBr3
so for 1) I converted 20g P4 to gPBr3
and 2.93gBr2 to gBPr3.
2) Limiting reagant will be Br2 cause its the lowest number
The maximum mass will be the outcome of conversions of both P4 and Br2 to PBr3 and add them together?
Not so.
mols P4 = 20/molar mass P4 = ?
mols Pbr3 = ?mol P4 x (4 mols PBr3/1 mol P4) = ?
For grams Br2, it isn't 2.93.
mass Br2 = volume x density = 50 mL x 2.93 g/mL = ?
Then mols Br2 = g/molar mass = ?
Convert mols Br2, as above, to mols PBr3 . The smaller number will be the correct value to chose for mols PBr3 produced and the reagent (P4 or Br2) creating that smaller value will be the limiting regent.
Then mass PBr3 = smaller mols PBr3 x molar mass PBr3. .
i want to no bout what kind of chemical reaction it has
To calculate the mass of the product (PBr3) formed, we need to determine the limiting reactant and then use stoichiometry.
1) Calculating the mass of product for each reactant:
- Phosphorus (P4):
The molar mass of P4 is 4 * 31.0 g/mol = 124.0 g/mol.
Since the reaction stoichiometry is 1:4 (P4: PBr3), the theoretical yield of PBr3 from 20 g of P4 is:
20 g P4 * (1 mol PBr3 / 124.0 g P4) * (4 mol PBr3 / 1 mol P4) * (270.0 g PBr3 / 1 mol PBr3) = 172.2 g PBr3.
- Bromine (Br2):
Density of Br2 = 2.93 g/mL.
Given volume of bromine = 50 mL.
The mass of bromine used is:
50 mL * (2.93 g/mL) = 146.5 g.
2) Determining the limiting reactant and the maximum mass of PBr3:
Since the stoichiometry is 1:6 (P4:Br2) and we have 20 g P4 and 146.5 g Br2, we need to compare the number of moles of each reactant.
- Phosphorus (P4):
20 g P4 * (1 mol P4 / 124.0 g P4) = 0.161 mol P4.
- Bromine (Br2):
146.5 g Br2 * (1 mol Br2 / 159.8 g Br2) = 0.916 mol Br2.
From the stoichiometry, we can see that we need 6 moles of Br2 for every mole of P4. Therefore, for the given reaction, 0.161 mol P4 would require (0.161 mol P4) * (6 mol Br2 / 1 mol P4) = 0.966 mol Br2.
Since the actual amount of Br2 used (0.916 mol) is less than the required amount (0.966 mol), Br2 is the limiting reactant. Therefore, the maximum mass of PBr3 that can be formed is determined by the amount of Br2.
Using the stoichiometry, we can calculate the maximum mass of PBr3:
0.916 mol Br2 * (4 mol PBr3 / 6 mol Br2) * (270.0 g PBr3 / 1 mol PBr3) = 155.7 g PBr3.
So the limiting reagent is Br2, and the maximum mass of PBr3 that can be formed is 155.7 grams.
To calculate the mass of product in grams for each of the reactants, you need to use stoichiometry, which is the quantitative relationship between the reactants and products in a chemical reaction.
1) Let's start with converting 20 grams of P4 to grams of PBr3:
- First, we need to determine the molar mass of P4 and PBr3.
- The molar mass of P4 is 123.88 g/mol (phosphorus has a molar mass of 30.97 g/mol).
- The molar mass of PBr3 is 270.62 g/mol (phosphorus has a molar mass of 30.97 g/mol, and bromine has a molar mass of 79.90 g/mol).
Now, we can set up the following stoichiometric ratio: 4 moles of PBr3 / 1 mole of P4. Knowing this ratio, we can find the moles of PBr3.
First, find the number of moles of P4:
moles of P4 = mass of P4 / molar mass of P4
moles of P4 = 20 g / 123.88 g/mol ≈ 0.161 moles of P4
Now, use the stoichiometric ratio to find the moles of PBr3:
moles of PBr3 = moles of P4 * (4 moles PBr3 / 1 mole P4)
moles of PBr3 = 0.161 moles P4 * (4 moles PBr3 / 1 mole P4) = 0.644 moles of PBr3
Finally, convert the moles of PBr3 to grams:
mass of PBr3 = moles of PBr3 * molar mass of PBr3
mass of PBr3 = 0.644 moles PBr3 * 270.62 g/mol ≈ 174.55 g
Therefore, the mass of PBr3 produced from 20 grams of P4 is approximately 174.55 grams.
2) To determine the limiting reactant, you compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces the lowest amount of product is the limiting reactant.
a) P4: As we calculated earlier, we have 0.161 moles of P4.
b) Br2: To calculate the moles of Br2, we need to convert the volume given in milliliters to grams using the density:
mass of Br2 = volume of Br2 * density of Br2
mass of Br2 = 50 mL * 2.93 g/mL = 146.5 g
Now, convert the mass of Br2 to moles using its molar mass:
moles of Br2 = mass of Br2 / molar mass of Br2
moles of Br2 = 146.5 g / 159.80 g/mol = 0.916 moles of Br2
To find the limiting reactant, we compare the moles of each reactant with their stoichiometric ratio in the balanced equation. From the balanced equation, we know that the ratio of moles of P4 to moles of Br2 is 1:6. So, for every mole of P4, we need 6 moles of Br2 to react completely.
Considering this ratio, we can compare the moles of P4 and Br2:
Moles of P4 / Stoichiometric ratio = 0.161 moles P4 / 1 = 0.161 moles PBr3
Moles of Br2 / Stoichiometric ratio = 0.916 moles Br2 / 6 = 0.153 moles PBr3
Comparing the moles, we can see that 0.153 moles of PBr3 is the lower amount produced. Therefore, Br2 is the limiting reactant.
To calculate the maximum mass of PBr3, we use the moles of the limiting reactant (Br2) and convert it to grams using its molar mass:
mass of PBr3 = moles of PBr3 * molar mass of PBr3
mass of PBr3 = 0.153 moles PBr3 * 270.62 g/mol ≈ 41.44 g
So, the maximum mass of PBr3 that can be obtained is approximately 41.44 grams when Br2 is the limiting reactant.