2 NaHCO_3 (s) rightarrow Na_2 CO_3 (s) + CO_2 (g) + H_2 O (g) NaHCO_3 (s) (bakin

Question

2 NaHCO_3 (s) rightarrow Na_2 CO_3 (s) + CO_2 (g) + H_2 O (g) NaHCO_3 (s) (baking soda) decomposes upon heating to product: Na_2 CO_3 (s) and two gaseous products. as shown by the equation above. (a) A student claims that the reaction is an oxidation-reduction reaction because the oxidation number of carbon changes. Do you agree with the claim? In your answer include the oxidation number of carbon in each of the three carbon-containing species in the reaction. The student conducts an experiment to determine the composition of a mixture of NaHCO_3 (molar mass 84.01 g/mol) and Na_2 CO_3 (molar mass 105.99 g/mol). The student places a sample of the mixture into a pre-weighed test tube that is attached to a container that bolds a drying agent The student heats, the test tube strongly with a Bunsen burner for 10 minutes, during which time all of the water produced by the reaction is captured by the drying agent. The following table shows the data the student recorded during the experiment. (b) Calculate the number of moles of NaHCO_3 (s) present in the mixture in the test tube before the reaction was initiated. (c) Determine the mass percent of NaHCO_3 (s) in the mixture. (d) If the student spills some of the mixture out of the text tube after weighing the test tube and mixture and before beating, bow would this error affect the mass percent of NaHCO_3 calculated in part (c) ? Justify your answer. When a sample of purr Na_2 CO_3 is placed in distilled water, the student observes that the pH of the solution increases significantly. This process is represented by the balanced net-ionic equation shown below. CO_3^2- (aq) + H_2 O (l) rightarrow HCO_3 (aq) + OH^- (aq) (e) The student prepares a 0.10 M Na_2 CO_3 (aq) solution and measures the pH of the solution to be 11.65. (i) Calculate [OH^-] in the Na_2 CO_3 (aq) solution. (ii) Write the expression for K_b for the carbonate ion. (iii) Calculate the value of K_b for the carbonate ion. The student adds some 1.0 M Sr (NO_3)_2 (aq) to the 0.10 M Na_2 CO_3 (aq) and observes the formation of a precipitate. (f) Write the balanced net-ionic equation for die reaction between Sr (NO_3)_2 and Na_2 CO_3, that produces the precipitate.

Explanation / Answer

2a) The reaction is not an oxidation-reduction reaction because the oxidation number of carbon is + 4 in both the compounds as can be shown below:

NaHCO3

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

The compound is electrically neutral; hence the overall charge on the compound is zero.

Let the oxidation number of C be x. Therefore,

1*(+1) + 1*(+1) + 1*x + 3*(-2) = 0

===> 2 + x – 6 = 0

===> x – 4 = 0

===> x = 4

The oxidation number of carbon is +4.

Na2CO3

Oxidation number of Na = +1

Oxidation number of O = -2

The compound is electrically neutral; hence the overall charge on the compound is zero.

Let the oxidation number of C be x. Therefore,

2*(+1) + 1*x + 3*(-2) = 0

===> 2 + x – 6 = 0

===> x – 4 = 0

===> x = 4

The oxidation number of carbon is +4.

Since both the compounds have carbon in the same oxidation state, the reaction is not a redox reaction. Infact, the reaction is a decomposition reaction.

b) The reaction taking place is

2 NaHCO3 (s) -----> Na2CO3 (s) + CO2 (g) + H2O (g)

The loss in mass of the mixture is due to the loss of CO2 and H2O gases due to the decomposition of NaHCO3.

As per the balanced equation above,

2 moles NaHCO3 = 1 mole H2O

Mass of sample mixture = (17.648 – 15.825) g = 1.823 g.

The drying agent will absorb the H2O gas formed and hence, the increase in mass of the drying agent.

Mass of H2O absorbed = (2.303 – 2.134) g = 0.169 g.

Molar mass of H2O = (2*1.008 + 1*15.9994) g mol-1 = 18.0154 g/mol.

Moles of H2O produced = (0.169 g)/(18.0154 g/mol) = 9.38*10-3 mole.

Moles of NaHCO3 present in the mixture = (9.38*10-3 mole H2O)*(2 moles NaHCO3/1 mole H2O) = 0.01876 mole.

Therefore, the number of moles of NaHCO3 present in the sample mixture = 0.01876 mole (ans).

c) Mass of NaHCO3 in the mixture = (0.01876 mole)*(84.01 g/mol) = 1.576 g.

Mass percent of NaHCO3 in the mixture = (1.576 g)/(1.823 g)*100 = 86.451% (ans).

d) The student spilled some of the mixture before heating the test tube. The amount of NaHCO3 in the mixture will be lower and hence the calculated mass percent of NaHCO3 in the mixture will be lower (ans).

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