1) Starting with 75.00 g of 1-propanol and 75.00 g of PCl 3. Calculate the theoretical yield of 1-chloropropane.
3CH3CH2CH2OH + PCl 3 --> 3CH3CH2CH2Cl + H3PO3 Answer in terms of grams.
the answer that i got is 98.12, is this right and how many sig figs should it be?
2) What is the concentration of a solution formed when 15.0 mL of 6.00 M HCl are diluted with 25.0 mL of water? Answer in terms of moles/L.
the answer that i got is 3.60, is this right and how many sig figs should it be?
3) What is the volume of a 3.72 M Na3PO4 solution needed to prepare 2.50 x 102 mL of a 1.43 M Na+ solution? Answer in terms of mL.
the answer that i got is 96.1, is this right and how many sig figs should it be?
4) What is the concentration of an HI solution if 75.0 mL of the acid required 35.21 mL of 0.1894 M KOH to reach the endpoint in the titration? Answer in terms of moles/L.
the answer that i got is 0.08892 is this right, and how many sig figs should it have?
I used 60.096 for the molar mass of the alcohol and 78 .54 for the molar mass of the 1-chloropropane and I obtained 98.01817. You are allowed 4 s.f. so I would round my answer to 98.02 grams.
#2 is 3.60 and you are allowed 3 s.f.
#3. I don't think so.
You want moles Na = 1.43M x 0.250 = ?
moles Na3PO4 = moles Na/3 = ?
M Na3PO4 = moles Na3PO4/L Na3PO4 = about 32 mL. You are allowed 3 s.f.
#4 is ok. Are you having trouble with s.f.? Here is a link that should help.
http://www.chemteam.info/SigFigs/SigFigs.html
are you sure about number 4? shouldn't it have 3 significant digits because of the 75.0 ml ?
1) To calculate the theoretical yield of 1-chloropropane, you need to determine the limiting reactant. To do this, convert both masses of 1-propanol and PCl3 to moles by dividing by their respective molar masses.
1-propanol (C3H8O):
Molar mass = 3(12.01 g/mol) + 8(1.01 g/mol) + 16.00 g/mol = 60.09 g/mol
75.00 g / 60.09 g/mol = 1.249 mol
PCl3:
Molar mass = 1(31.00 g/mol) + 3(35.45 g/mol) = 137.33 g/mol
75.00 g / 137.33 g/mol = 0.546 mol
The balanced equation shows a 1:1 ratio between 1-propanol and 1-chloropropane, so the limiting reactant is PCl3. Therefore, the moles of 1-chloropropane formed will be equal to the moles of PCl3 used.
The molar mass of 1-chloropropane (C3H7Cl) is:
3(12.01 g/mol) + 7(1.01 g/mol) + 35.45 g/mol = 74.12 g/mol
0.546 mol of PCl3 will produce 0.546 mol of 1-chloropropane.
Mass of 1-chloropropane = moles × molar mass
Mass = 0.546 mol × 74.12 g/mol = 40.45 g
So, the theoretical yield of 1-chloropropane is 40.45 grams.
Considering significant figures, the given masses of the reactants include three significant figures. According to multiplication and division rules, the answer should also be reported with three significant figures. Thus, the theoretical yield of 1-chloropropane should be 40.5 grams.
2) To calculate the concentration of the diluted solution, you can use the formula:
C1V1 = C2V2
where C1 and V1 are the initial concentration and volume of the concentrated solution, respectively, and C2 and V2 are the final concentration and volume of the diluted solution, respectively.
Given:
C1 = 6.00 M
V1 = 15.0 mL
C2 (unknown)
V2 = 25.0 mL
Rearrange the formula to solve for C2:
C2 = (C1V1) / V2
C2 = (6.00 M × 15.0 mL) / 25.0 mL
C2 = 3.6 M
So, the concentration of the solution formed is 3.6 moles/L.
Since the given concentration has three significant figures, the answer should also be reported with three significant figures. Thus, the concentration is 3.60 moles/L.
3) To calculate the volume of the Na3PO4 solution, you can use the formula:
C1V1 = C2V2
where C1 and V1 are the initial concentration and volume of the Na3PO4 solution, respectively, and C2 and V2 are the final concentration and volume of the Na+ solution, respectively.
Given:
C1 = 3.72 M
V1 (unknown)
C2 = 1.43 M
V2 = 2.50 × 102 mL = 250 mL
Rearrange the formula to solve for V1:
V1 = (C2V2) / C1
V1 = (1.43 M × 250 mL) / 3.72 M
V1 = 96.1 mL
So, the volume of the 3.72 M Na3PO4 solution needed to prepare 250 mL of a 1.43 M Na+ solution is 96.1 mL.
Considering significant figures, the given volumes have three significant figures, and the given concentration has four significant figures. According to division and multiplication rules, the answer should be reported with three significant figures. Thus, the volume should be 96.1 mL.
4) To calculate the concentration of the HI solution, use the stoichiometry of the balanced equation:
HI + KOH -> KI + H2O
The molar ratio between HI and KOH is 1:1 from the balanced equation.
Given:
Volume of HI solution = 75.0 mL = 0.075 L
Volume of KOH solution = 35.21 mL = 0.03521 L
Concentration of KOH = 0.1894 M
Use the formula:
Molarity1 × Volume1 = Molarity2 × Volume2
Molarity2 = (Molarity1 × Volume1) / Volume2
Molarity2 = (0.1894 M × 0.03521 L) / 0.075 L
Molarity2 = 0.08892 M
So, the concentration of the HI solution is 0.08892 moles/L.
Considering significant figures, the given volumes have three significant figures, and the given concentration has four significant figures. According to division and multiplication rules, the answer should be reported with four significant figures. Thus, the concentration should be 0.08892 moles/L.