For questions 13-14, use the information provided below from the experiment Iden
ID: 925974 • Letter: F
Question
For questions 13-14, use the information provided below from the experiment Identification of an Unknown. The molecular masses and reactions are provided for you.
NaHCO3 KHCO3 Na2CO3 K2CO3 NaCl KCl
84.01 g/mol 100.12 g/mol 105.99 g/mol 138.21 g/mol 58.44 g/mol 74.55g/mol
2 NaHCO3 (s) à Na2CO3 (s) + H2O (g) + CO2 (g) (Reaction 1)
Na2CO3 (s) +2HCl (aq) à 2NaCl (s) + H2O (l) + CO2 (g) (Reaction 2)
13. Brad starts with 0.652 g of unknown. How much mass should he have after the first heating step if his unknown
was NaHCO3?
A. 0.411 g Na2CO3
B. 0.326 g Na2CO3
C. 0.822 g Na2CO3
D. 0.00402 g Na2CO3
14. If Brad has a final mass of 0.270 g at the end of the experiment (after the 2nd heating), what is his percent yield
of the final salt? (Assuming he starts with 0.652 g of NaHCO3).
A. 40%
B. 60%
C. 80%
D. 86%
15. What is the concentration of chloride ions in 50.0 mL of a 5.25 M solution of barium chloride?
A. 0.525 M Cl-
B. 1.05 M Cl-
C. 5.25 M Cl-
D. 10.5 M Cl-
Aqueous Solutions
16. Which of the following represents a correct precipitation reaction between sodium iodide and lead(IV) sulfate?
A. PbSO4 (aq) + NaI(aq) à PbI (aq) + NaSO4 (aq)
B. PbSO4 (aq) + NaI(aq) à PbI (s) + NaSO4 (aq)
C. Pb(SO4)2(aq) + NaI (aq) à PbI2 (s) + Na2SO4 (aq)
D. Pb(SO4)2(aq) + NaI (aq) à PbI2 (aq) + Na2SO4 (aq)
17. How many of the following salts would not form a precipitate in an aqueous solution?
A. one
B. two
C. three
D. four
18. What type of reaction is the following? CaCO3 + heat à CaO + CO2
A. Combination
B. Decomposition
C. Single Replacement
D. Combustion
19. Calculate the amount of copper recovered from an 85% yield if the theoretical yield is 0.25 g
A. 21 g
B. 3.4 g
C. 0.21 g
D. 0.34 g
20. Which of the following chemical species is the oxidizing agent in the reaction below?
Cu (s) + 4 HNO3 (aq) Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)
A. Cu (s)
B. HNO3 (aq)
C. Cu(NO3)2 (aq)
D. NO2 (g)
Explanation / Answer
13)
the first reaction is
2 NaHC03 ---> Na2C03 + H20 + C02
we know that
moles = mass / molar mass
so
moles of NaHC03 taken = 0.652 / 84
moles of NaHC03 taken = 7.76 x 10-3
now
from the first reaction
we can see that
moles of Na2C03 formed = 0.5 x moles of NaHC03 taken
so
moles of Na2C03 = 0.5 x 7.76 x 10-3 = 3.88 x 10-3
now
mass = moles x molar mass
so
mass of Na2C03 = 3.88 x 10-3 x 105.99
mass of Na2C03 = 0.411 g
so
A) 0.411 g Na2C03
14)
now
Na2C03 + 2 HCl ---> 2 NaCl + H20 + C02
we can see that
moles of NaCl = 2 x moles of Na2C03
moles of NaCl = 2 x 3.88 x 10-3 = 7.66 x 10-3
now
mass = moles x molar mass
so
mass of NaCl = 7.66 x 10-3 x 58.44 = 0.44765
so
0.44765 grams of NaCl should be formed
but
given only 0.27 grams is formed
now
% yield = actual x 100 / theoretical
% yield = 0.27 x 100 / 0.44765
% yield = 60.3
so
B) 60 %
15)
BaCL2 ---> Ba+2 + 2Cl-
we can see that
[Cl-] = 2 x [BaCl2]
[Cl-] = 2 x 5.25
[Cl-] = 10.5
D) 10.5 M Cl-
16)
C)
18) given reaction is
CaC03 + heat ---> CaO + C02
it is decomposition reaction
the answer is B
19) we know that
% yield = actual x 100 / theoretical
85 = actual x 100 / 0.25
actual = 0.2125
so
C) 0.21 g
20)
we know that
oxidizing agent undergoes reduction
in the given reaction
nitrogen undergoes reduction
so
HN03 is the oxidizing agent
so
B) HN03 (aq)