Lap work 1 - Lecture notes 1-7 PDF

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Laboratory work #1 Limiting Reagents Problem Distributed:

Due:

Name :- Malak Munef Mohammed Abdullah Al-hammdi

Objective: To practice with experiments involving limiting reagents and the test your knowledge to determine the concentration of an unknown solution. Assignment: Problem #0 In this problem we’ll consider the reaction NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s) Note that all of the aqueous species in the above reaction exists as ions in solution: NaCl(aq) exists as Na+ and ClAgNO3 exists as Ag+ and NO3NaNO3 exists as Na+ and NO3-

AgCl(s) is a white solid.

Preform answers for the following questions :1. The solution labeled “1.00g NaCl” contains 1.00g of NaCl dissolved in water. Use the solution viewer to determine the number of grams of Na+ and Cl- in the solution, and confirm that these add to 1.00g. Na ion and Cl ion is 0.170529

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Therefore we will convert the mol into gram by multiplying each with respective atomic mass…… Na ion = 0.170529mol x 22.990 g/mol Na ion = 3.9205 gram Cl ion = 0.170529 mol x 35.453 g/mol Cl ion =0.60458 gram Recheck answers by changing to gram unit

2. The solution labeled “1.00g AgNO3” contains 1.00g of solid AgNO3. Add 100ml of water to this solution. Use the solution viewer to determine the number of grams of Ag + and NO3- in the solution.

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-For Ag and No3 we will preform the same calculations method and compare the value with simulation…

3. Now add the 1.00g of solid AgNO3 to the 1.00g solution of NaCl. Write down the mass of each species in solution and the mass of solid AgCl formed. Confirm that the amounts of these species are consistent with what you say in parts (2) and (3). -

For this part the mass of the species Is shown as follows :-

The AgCl form was around 0.843704 gram where in term of mol unit would be 0.00588685 mol of AgCL. This value will be used for comparison in the next calculation :Note that the Na ion mass is the same as the pervious, but the Cl ion has been reduced to around half of the original mass and that’s because the other half of the Cl is not in aqueous form but have been attached to Ag ion to form AgCl precipitate. It can be concluded that with 1 gram of NaCl solution react with 1 gram of AgNo3 aqueous solution The limiting reactant AgNo3. Compare withtheoretical calculation :NaCl(aq) + AgNO3(aq)NaNO3(aq) + AgCl(s)

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Based on the simulation experiment,1g of NaCl react with 1g AgNo3 to form NaNo3 aqueous solution and AgCl precipitation, assuming the reaction is complete reaction Mol of NaCl = 1g of NaCl/(22.990+ 35. 453)g/mol = 1/ 58.443 0. 0171 mol of NaCl Mol of AgNo3 = 1g of AgNo3 /( 107.868 +14.007+(15.999x3)) = 1/ 169.854 0.00589 mol of AgNO3 Baced on the Stoichiometric equation, 1mol of NaCl is react with 1mol AgNo3 to form 1 mol of AgCl Based on the comparison of NaCl and AgNo3 in terms of amount substance ( mol), NaCl is higher making it to be the excess reactant Substance Stoichiometrc ratio Mol ratio

NaCl 1 0.0171

AgNO3 1 0.00589

As seen in the table the smallest mol value is 0.00589 Makes the AgNO3 becomes the Limiting Reactant therefore the reaction will stop whem all mol of AgNO3 is used up and 0.00589 of AgNO3 precipitate will be formed. Compare the value with simulation

With this comparison we conclude that the value is the same when comparing the theoretical calculation with the simulation. This proves that the above statement is true where NaCl is the excess reagent and the AgNO3 is the limiting reactant.

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Problem #1

The solution labeled “Solution 1” in the virtual lab stockroom contains 2.00 grams of Sodium Chloride. 1.

How many grams of Silver Nitrate must be added to the solution to completely react with Sodium Chloride according to the reaction below: NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s)

Note that all of the aqueous species in the above reaction exists as ions in solution: NaCl(aq) exists as Na+ and ClAgNO3 exists as Ag+ and NO3NaNO3 exists as Na+ and NO3-

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AgCl(s) is a white solid.

We have 2 g of NaCl so moles of the salt is = 2.0 g / 58.4428 g/mol = 0.0342 moles

These will react with 0.0342 moles of AgNO3 according to our equation Mass of AgNO3 in these moles = 0.0342 mole * 169.8731 g/mol = 5.78 g of AgNO3 2.

Use the virtual lab to add that amount of Silver Nitrate to the solution. Check to make sure the reaction was complete, by making sure the amount of Ag+ and Cl- in the solution are both less than 0.01g. Problem #2

The solution labeled “Solution 2” in the virtual lab stockroom contains 3.00 grams of AgNO 3. When NaCl is added to the solution, the following reaction occurs: NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s) 1.

If excess NaCl is added to the solution, how many grams of AgCl(s) will be formed?

2.

Use the virtual lab to check your answer to part (a). Explain the laboratory procedure you used to perform this check. Limiting Reagents Problem 2

Continued from Limiting reagents problem 1: The solution labeled “Solution 3” in the virtual lab stockroom contains an unknown amount of AgNO 3. When NaCl is added to the solution, the following reaction occurs: NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s) Design and perform an experiment to determine the mass of AgNO3: solution is 3.37grams of of AgNO3 . 1.

Add solid NaCl to the solution in small increments (0.1 g). You should start to see AgCl(s) form in the flask. Continue to add NaCl until the amount of AgCl no longer increases when you add the salt. This is what we mean by adding “excess” NaCl.

2.

Determine the mass of AgCl that was formed.

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3.

From the mass in part 2, determine the amount of AgNO3 that must have been present in the initial solution.

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