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 2^nd 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)

Answer 1

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)