2-4 410 81%. 7:16 AM X Question Chemistry 101 63 Name: Tear this page out of you

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2-4

410 81%. 7:16 AM X Question Chemistry 101 63 Name: Tear this page out of your laboratory manual and submitted to your TA with your answers B. Post-Lab Assessment Questions: (20 pts) . A reaction between sodium hydroxide and hydrochloric acid produces sodium chloride and water a) (2 pts) Write the balanced chemical equation for this reaction. b) (3 pts) If 22.85 g of sodium hydroxide reacts completely with 20.82 g of hydrochloric acid to form 10.29 g of water, what mass of sodium chloride is formed during the reaction? 2. (8 pts) When a solution of barium nitrate and a solution of copper (II) sulfate are mixed, a chemical reaction produces solid barium sulfate and an aqueous solution of copper (II) nitrate. An unknown amount of barium nitrate is dissolved in 120.0 g of water and 8.15 g of copper (II) sulfate is dissolved in 75.0 g of water. When these solutions are mixed 1.076 g of white solid is isolated as a precipitate by vacuum filtration. The filtrate is found to have a mass of 204.44 g. What mass of barium nitrate was dissolved in the water and used for this reaction? You must show balanced chemical equation and your calculations to receive credit for this problem. 3. (3 pts) When a wooden log burns it is essentially undergoing a combustion reaction. Using this information and what you know about mass conservation, explain why the ashes that form when a log burns have a lower mass than the log had before it started burning (4 pts) Consider the following chemical reaction: CuSO4 (s) + heat Cu0 (s) + SO3 (g) Design an experiment in which you use this chemical reaction to test the law of mass conservation.

Explanation / Answer

Ans. #2. Balanced reaction:           

CuSO4(aq) + Ba(NO3)2(aq) -------> BaSO4(s) + Cu(NO3)2 (aq)

Moles of CuSO4 taken = Mass / Molar mass = 8.15 g / (159.6096 g/mol) = 0.051062 mol

Moles of BaSO4 precipitate formed = 1.076 g/ (233.39060 g/ mol) = 0.004610 mol

# According to the stoichiometry of balanced reaction, 1 mol BaSO4 is precipitate is formed by 1 mol CuSO4.

Note that the moles of BaSO4 is around 11 times less than that of the CuSO4. It means CuSO4 was the reagent in excess whereas Ba(NO3)2 was the limiting reactant.

So, using stoichiometry of balanced reaction-

            Moles of Ba(NO3)2 taken = 0.004610 mol = Moles of BaSO4 formed

Now,

            Mass of Ba(NO3)2 taken = Moles x Molar mass

                                                = 0.004610 g x (261.33688 g/ mol)

                                                = 1.205 g

#3. Ash is the un-combustible inorganic content in the log. For example, it may be silica or oxides of minerals like CaO, Na2O, MgO, Fe2O3, etc. The inorganic minerals constitute very small portion (say, around 1% of total mass) of the whole mass of an organism or log.

The loss in mass during burning is due to combustion of organic matter like carbohydrates, lipids, etc. Since organic matter is chiefly constituted of C, H and O, burning those produces CO2 and H2O- which are lost as gases.

That is-

Mass of ash = (Mass of log) - (Mass of organic matter lost in form of CO2 and H2O)

Therefore, mass of ash (oxides of inorganic minerals) is always less than that of the whole log because of loss of bulk of organic matter in form of CO2 and H2O during combustion.

#4. Step 1: Take X gram of CuSO4 in a crucible (for eliminating complexities in calculation, it’s assumed that the mass of crucible is constant throughout the process).

Mass of CuSO4 taken = X g

Step 2: Heat the sample to get a constant mass.

            Mass of residue (CuS) after heating = Y gram

Calculate the moles of CuS remaining in the sample after heating by using the formula-

            Number of moles of CuS = Mass/ Molar mass = M mol

Step 3: According to the stoichiometry of balanced reaction, formation of 1 mol CuS is accompanied with formation of 1 mol SO3

So,

            Moles of SO3 produced = M mol = Moles of CuS residue

Calculate the mass of SO3 lost by using the formula-

            Mass of SO3 lost = M mole x Molar mass

Let the mass of SO3 lost = Z gram

Step 4: Validation of law of mass conservation

Total mass of reactants = Total mass of products

            Or, X g = Y g + Z g             - equation 1

If you perform the experiment correctly, equation 1 would be experimentally valid that further proves the law of mass conservation.

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