Break your team up. One will work on Goal #1, one will work on Goal #2 and the o
ID: 564901 • Letter: B
Question
Break your team up. One will work on Goal #1, one will work on Goal #2 and the other members will work on Goal #3. PRIOR TO CLASS CALCULATIONS: 1. Calculate the masses of Ca(NO,)2 4H Ot and tN0e required to make 10.0 g of Ca(IO3)2/s). Note: Calculate the amount of CeNOjr. 4H2O(s) needed so that the Ca2+ is in excess of the iodate ion concentration by 20% (i.e., colculate a weight of calcium nitrate that is 20% higher than the minimum requird to produce 10.0 g of calcium iodate). 2. Calculate the mass of Ca(NO3)2 4H20 required to generate a 0.1XX M Ca2 solution in a 100 mL volumetric flask. it is written as 0.1xX M because you don't know the actual concentration of calcium until you weigh the solid and generate the solution, but here is what you do know: (a.) it should be close to 0.1 M; (b.) the final concentration should be known out to 3 significant figures; and (c.) you should use the actually generated concentration for all of your subsequent calculations (not just 0.1 M). 3. Document your proposed dilution scheme on the laboratory data sheet for the serial dilution on page 6 Thoroughly document all of your calculations and dilution scheme on page 13Explanation / Answer
Ans.
#1. Moles of Ca(IO3)2 to be made = Mass / Molar mass
= 10.0 g / (389.88334 g/ mol)
= 0.0256 mol
# 1 mol Ca(IO3)2 consists of 1 mol Ca2+ and 2 moles IO3-.
# Required amount of Ca(NO3)2.4H2O:
1 mol Ca(NO3)2.4H2O yields 1 mol Ca2+.
So,
Required moles of Ca(NO3)2.4H2O = required moles of Ca2+ = 0.0256 mol
Required mass of Ca(NO3)2.4H2O = Required moles x Molar mass
= 0.0256 mol x (236.149 g/ mol)
= 6.0569 g
# Required amount of KIO3:
1 mol KIO3 yields 1 mol IO3-.
Required moles of IO3- = 2 x Moles of Ca(IO3)2 required = 2 x 0.0256 mol = 0.0513 mol
So,
Theoretical required moles of KIO3 = required moles of IO3- = 0.0513 mol
Theoretical required mass of KIO3 = Required moles x Molar mass
= 0.0513 mol x (214.00 g/ mol)
= 10.9776 g
# Actual required mass of KIO3=
Theoretical required mass + (20% of Theoretical required mass)
= 10.9776 g + (20% 0f 10.9776 g)
= 13.1732 g
#2. Let the required concertation of Ca2+ = 0.123 M
Given, volume of solution = 100.0 mL = 0.100 L
Now,
Required moles of Ca2+ = Molarity x Volume of solution in liters
= 0.123 M x 0.100 L
= 0.0123 mol
# 1 moles Ca(NO3)2.4H2O yields 1 mol Ca2+.
So,
Required moles of Ca(NO3)2.4H2O = required moles of Ca2+ = 00.0123 mol
Required mass of Ca(NO3)2.4H2O = Required moles x Molar mass
= 0.0123 mol x (236.149 g/ mol)
= 2.9046 g
Note: If you need other calculations like 0.101 M, follow the above steps to get the required mass.
#2”. Moles of reagent A taken = Molarity x Volume of solution in liters
= 0.100 M x 0.010 L
= 0.001 mol
# In the balanced reaction, 1 mol A is neutralized by 3 mol B.
So,
Required moles of B = 3 x Moles of A = 3 x 0.001 mol = 0.003 mol
# Given, volume of solution B consumed = 33.24 mL = 0.03324 L
Now,
Molarity of solution B = Moles of B/ Volume of solution B in liters
= 0.003 mol / 0.03324 L
= 0.0903 M
#3. According to the stoichiometry of balanced reaction, 1 mol PbI2 yields 1 mol Pb2+ and 2 mol I- upon dissociation. That is, always 2 moles of I- per 1 mol Pb2+ in the solution.
So,
[Pb2+] = (½) x [I-] = (½) x 0.0032 M = 0.0016 M
#4. According to the stoichiometry of balanced reaction, 1 mol PbBr2 yields 1 mol Pb2+ and 2 mol Br- upon dissociation. That is, always 2 moles of Br- per 1 mol Pb2+ in the solution.
So,
[Pb2+] = (½) x [Br-] = (½) x 0.0012 M = 0.0006 M
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