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Hey all! Thanks for the help in advance.

I thumbs up all correct answers! :) Have a nice day!

Sheet #4 To Be Turned In-Show All Work For Full Credit! Name Part 5 Practice: Solution and Dilution Calculations, continued 5. What is the molarity of a KMnO, solution that was prepared by mixing 14.2 mL of a Lab Day & Time 0.688 M KMnO, solution with enough water to prepare 50.0 mL of solution? 6. How many mL of 0.450 M KNO, must be added to a volumetric flask and diluted with enough water to make 250. mL of a 0.00950 M KNOs solution? 7. What is the concentration of Fe3 and the concentration of NO3 present in the solution that results when 30.0 mL of 1.75 M Fe(NO;)s are mixed with 45.0 mL of 1.00 M HCI? 8. How much water would have to be added to 135 mL of 3.00 M H2SO4 to prepare a 1.25 M H2SO, solution? (Hint: What must the volume of the final solution be?) CEM 122 Review of Chemical and Mathematical Concepts 14

Explanation / Answer

5) This is the case of dilution, here we use

M1V1 =M2V2

0.688 *14.2 = M * 50

Molarity of KMnO4 = 0.195M

6) Using the above formula volume of KNO3 can be calculated as

M1V1 =M2V2

0.45* V =0.0095* 250

V= 5.27ml

7) When Fe(NO3)3 mixed with HCl following ions are present in resulting solution

Fe(NO3)3 + HCl----> Fe3+ + 3 NO-3 + H+ + Cl-

total volume of solution= 30+ 45= 75ml

Using M1V1=M2V2

[Fe3+] = 1.75*30/75 = 0.7M

Conc. Of NO3- = 3 times of [Fe3+] =3 *0.7 =2.1M

As 1mol of Fe(NO3)3 contain one mol Fe3+ and 3 mol of NO3-

8) using M1V1 = M2V2

3 * 135 =1.25 *V

V Of solution = 324ml.

So, Vol of water added = 324-135 = 189ml

9) conc. Of Mg2+ = total moles of Mg2+/ total vol. Of solution

= (M1V1 +M2V2 )/ (V1+V2)

= (0.888*65 +0.475 *35 )/ (65+ 35)

= 0.74M

10) 2FeCl3 + Na2CO3 -----> Fe2(CO3)3 + 6 NaCl

Moles of FeCl3 = 15 *0.15 =2.25millimoles

Moles of Na2CO3 = 0.15 *20 = 3millimoles

According to stoichiometry, FeCl3 is limiting reagent

So formation of Fe2(CO3)3 depends on FeCl3

According to stoichiometry,

2 moles of FeCl3 forms 1mol of Fe2(CO3)3

So 2.25 millimole FeCl3 will produce 1.125millimole Fe2(CO3)3

Mass of 1.125millimole of Fe2(CO3)3 = moles* molar mass

= 1.125 * 291.7 /1000 g = 0.328 g

(Molar mass of Fe2(CO3)3 = 291.7g/mol)

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