Google Dich aTables.pe X1045 Exp7 AnalysisGaseo x 104 eriments/1045 Exp7 Analysi

Question

Google Dich aTables.pe X1045 Exp7 AnalysisGaseo x 104 eriments/1045 Exp7 AnalysisGaseousProduct.pdf m vdi G...my study lization x Pre-lab 1. 2. Read the Introduction and Experimental Procedure. List all the data that needs to be measured. You will need to weigh approximately 2.3 g of the ground garden lime. Write a detailed procedure on how to weigh it 3. Using the data in Example 1, perform the following calculations, showing all your work. Calculate the moles of carbon dioxide produced in Trial 1. Hint: Consider which gas law you will use for this calculation. Which measurements require unit conversion for the law? (See pages 2.4.) a. b. Calculate the mass of calcium carbonate in Trial 1. c. Verify that the 96 calcium carbonate in Trial 1 was calculated correctly d. Verify that the mean % calcium carbonate in sample 37 was calculated correctly e. Verify that the standard deviation was correctly reported f Verify that the relative percent error was correctly reported. 4. A pure sample of calcium carbonate prpduced 0.998 g CO, What was the mass of the calcum 5. How many ml of 3.0 M HCl are needed to completely react with 2.500 g of pure calcium carbonate used? carbonate? [Note: You will use less than this calculated volume of HCI for each trial. Think about the maximum volume of HCl you will need. To prevent waste, do not take more than what you need to perform the experiment] Post-lab 1. Include the data collected in the lab. 2. Write the balanced chemical equation that relates CO, and calcium carbonate. 3. For Trial 1: a. Show calculation for the pressure of the CO b. Show calculation for the percent by mass of calcium carbonate. 4. Calculate (show calculations): an nercent hy mass of calcium carbonate

Explanation / Answer

3) I need the data provided in Example #1 to answer the question.

4) Write the balanced chemical equation for the decomposition of calcium carbonate (CaCO3) as

CaCO3 (s) -----------> CaO (s) + CO2 (g)

As per the stoichiometric equation,

1 mole CaCO3 = 1 mole CO2.

Molar mass of CO2 = (1*12.011 + 2*15.999) g/mol = 44.009 g/mol.

Mole(s) of CO2 corresponding to 0.998 g CO2 = (0.998 g)/(44.009 g/mol) = 0.02268 mole.

As per the stoichiometric equation, mole(s) of CaCO3 decomposed to yield 0.02268 mole CO2 = 0.02268 mole.

Molar mass of CaCO3 = (1*40.078 + 1*12.011 + 3*15.999) g/mol = 100.086 g/mol.

Mass of CaCO3 corresponding to 0.02268 mole CO2 = (0.02268 mole)*(100.086 g/1 mole) = 2.2699 g 2.270 g (ans).

5) Write the balanced chemical equation for the reaction between CaCO3 and HCl as

CaCO3 (s) + 2 HCl (aq) ---------> CaCl2 (aq) + CO2 (aq) + H2O (g)

As per the stoichiometric equation,

1 mole CaCO3 = 2 moles HCl.

We have already noted the molar mass of CaCO3 as 100.086 g/mol.

Mole(s) of CaCO3 corresponding to 2.500 g CaCO3 = (2.500 g)/(100.086 g/mol) = 0.02498 mole.

Mole(s) of HCl required = (0.02498 mole CaCO3)*(2 moles HCl/1 mole CaCO3) = 0.04996 mole.

Volume of 3.0 M HCl required = (moles of HCl)/(molarity of HCl) = (0.04996 mole)/(3.0 M) = 0.01665 L = (0.01665 L)*(1000 mL/1 L) = 16.65 mL 16.7 mL (ans).

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