So I know that the dilution formula is M1V1=M2V2 but do not understand how to ap

Question

So I know that the dilution formula is M1V1=M2V2 but do not understand how to apply it here.

Step 2 of the procedure will require that you prepare one of the five standard solutions outlined in the table below. You will need to know the molarity of each solution in order to perform subsequent calculations. Use the dilution formula to complete the table below and then transfer these calculations to the Part I Data Table above this set of questions 5. 0 Concentration of FeSCN2] 0.00 0.200 M Fe(NO3)3 0.0020 M SCN- H20 (mL) Standard number (mL) (mL) 5.0 5.0 5,0 5.0 5.0 0.0 2.0 3.0 4.0 5.0 45.0 43.0 42.0 41.0 40.0

Explanation / Answer

The balanced chemical reaction is :

Fe+3       +      SCN-     = FeSCN+2

For each of the experiment: total volume of final solution is 50 mL

Solution 2:

5 mL of 0.200 M Fe(NO3)3 solution contains (5 *0. 200 )/ 1000 moles of Fe+3 = 1* 10-3 mole of Fe+3

2 mL of 0.002 M SCN- solution contains (2 *0. 002 )/ 1000 moles of SCN- = 4 * 10-6 mole of Fe+3

SCN- is the limiting reagent and according to reaction stoichiometry 4 * 10-6 mole of SCN- forms 4 * 10-6 mole FeSCN+2 and the volume of reaction mixture is 50 mL.

So, concentration of FeSCN+2 in solution is: (4 * 10-6 * 1000) / 50 M = 8* 10-5 M

Solution 3:

5 mL of 0.200 M Fe(NO3)3 solution contains (5 *0. 200 )/ 1000 moles of Fe+3 = 1* 10-3 mole of Fe+3

3 mL of 0.002 M SCN- solution contains (3 *0. 002 )/ 1000 moles of SCN- = 6 * 10-6 mole of Fe+3

SCN- is the limiting reagent and according to reaction stoichiometry 6 * 10-6 mole of SCN- forms 6 * 10-6 mole FeSCN+2 and the volume of reaction mixture is 50 mL.

So, concentration of FeSCN+2 in solution is: (6 * 10-6 * 1000) / 50 M = 1.2* 10-4 M

Solution 4:

5 mL of 0.200 M Fe(NO3)3 solution contains (5 *0. 200 )/ 1000 moles of Fe+3 = 1* 10-3 mole of Fe+3

4 mL of 0.002 M SCN- solution contains (4 *0. 002 )/ 1000 moles of SCN- = 8 * 10-6 mole of Fe+3

SCN- is the limiting reagent and according to reaction stoichiometry 8 * 10-6 mole of SCN- forms 8 * 10-6 mole FeSCN+2 and the volume of reaction mixture is 50 mL.

So, concentration of FeSCN+2 in solution is: (8 * 10-6 * 1000) / 50 M = 1.6* 10-5 M

Solution 5:

5 mL of 0.200 M Fe(NO3)3 solution contains (5 *0. 200 )/ 1000 moles of Fe+3 = 1* 10-3 mole of Fe+3

5 mL of 0.002 M SCN- solution contains (5 *0. 002 )/ 1000 moles of SCN- = 1 * 10-5 mole of Fe+3

SCN- is the limiting reagent and according to reaction stoichiometry 1 * 10-5 mole of SCN- forms 1 * 10-5 mole FeSCN+2 and the volume of reaction mixture is 50 mL.

So, concentration of FeSCN+2 in solution is: (1 * 10-5 * 1000) / 50 M = 2* 10-4 M

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