Report for Experiment 8 a. 43477N IDENTIFYING THE UNKNOWN s Recall, the number o

Question

Report for Experiment 8 a. 43477N IDENTIFYING THE UNKNOWN s Recall, the number of moles massmolar mass, so here moles of Naci mass Nacw585 g/mol). Similarly, if the unknown were NaHco, call itcompound A), the sample would contain this many this sample weight(84.0 g/mol) AND irit were Natco, call it compound B). it ouy contain many moles: sample weight/I060 gmol). Make these calculations and enter your values in this table: ,309 Possible Number of Mass (g) Number of Starting Molar Moles IF of NaCI Moles of Mole Namber Material Mass Unknown Lobtained NaCI Ratios Mole Ratio 1.002 NaHCO, 84.0 CIA Naco, 106.0 B: .This is the ratio of moles of Naci produced to moles of carbonate used. Remember that the whole number mole ratio for one of the two cabonates. the one you have in your unknown, will be close to 1 to 1 or 2 to 1. RESULTS 1. Your unknown is which carbonate? 2. Using your conclusion on identity and whole number mole ratios, write the balanced equation for the reaction of your unknown with 3. what specifie nerimental errors might cause your whole number mole ratios to differ from ts of the balanced equation obtained by the method? of the reaction, what small) climatic effect has your experiment had? Considering the products

Explanation / Answer

Ans. Part A: Balanced reaction: NaHCO3(s) + HCl(aq) ----------> NaCl(aq)+ H2O(l) + CO2(aq)

Stoichiometry: 1 mol NaHCO3 produces 1 mol NaCl

Therefore, if the molar ratio of product / reactant = 1, then the reactant is NaHCO3 because each mol of Na2CO3 produces one the moles of NaCl.

II. Balanced reaction: Na2CO3(s) + 2 HCl(aq) ----------> 2 NaCl(aq)+ H2O(l) + CO2(aq)

Stoichiometry: 1 mol Na2CO3 produces 2 mol NaCl.

Therefore, if the molar ratio of product / reactant = 2, then the reactant is Na2CO3 because each mol of Na2CO3 produces twice the moles of NaCl.

So, the determination of the product is bases on the stoichiometry of reaction I and II.

Part B. Coming to the data provided in table:

C/A ratio = moles of NaCl / moles of NaHCO3

                        0.109 / 0.99 = 0.11

Theoretically, you should have been getting a value near 1 or 2 depending on the unknown.

Now, Look carefully:

                        # You take 0.99 mol (= 83.16 g) NaHCO3

                        # You get 0.109 mol (= 6.36 g, more accurately 6.40 g)

If-

- your measurements were correct

- you added sufficient amount of HCl (i.e. HCl was present in excess)

- you allowed sufficient reaction time, and

-you processed rest of the experiment carefully, then

- you should have been getting around 0.99 mol NaCl.

However, you get only 1/10 of the expected amount of product.

The same has happened to unknown 2 also.

So, please review your data carefully. Your data are not consistent with the experimental values or some errors might have crept somewhere in the experiment.

For example: Consider a hypothetical example, similar to unknown 1 or 2

            Moles of unknown (1 or 2, I don’t known) = 0.099

            Moles of product obtained = 0.109

Molar ratio of product and reactant –

            Moles of product / moles of reactant = 0.099 mol / 0.109 = 0.908 = 1 (nearest whole number).

The product/ reactant or C/ (A or B) ratio of means that 1 mol product is formed by 1 mol reactant. It is seen when NaHCO3 is the reactant (see, reaction I). So, the reactant giving C/A ratio of 1 is NaHCO3.

Similarly, you should get C/B around 2.0   

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