A potassium dichromate is prepared by dissolving 0.3525 grams of K_2Cr_2O_7 is e

Question

A potassium dichromate is prepared by dissolving 0.3525 grams of K_2Cr_2O_7 is enough water to yield 250.0 mL of solution. A student then measures 10.00 mL of an unknown solution of Fe^2+ into an Erlenmeyer flask and dilutes with 90 mL of 1.0 M hydrochloric acid. He then titrates this sample with the solution prepared above. If 38.40 mL of the potassium dichromate solution is needed to reach the end point, what is the concentration of Fe^2+ in the original unknown solution? The balanced equation for the reaction of aqueous sulfuric acid with aqueous ammonia is 2NH_3(aq) + H_2SO_4 (aq) rightarrow (NH_4)_2SO_4(aq) What volume of 0.250 M sulfuric acid solution would be needed to react completely with 18.00 mL of 0.350 M ammonia solution? What mass of ammonium sulfate would be formed by this reaction?

Explanation / Answer

Ans. 3. Balance equation: Cr2O72- + 6 Fe2+ + 14 H+ -------> 2 Cr3+ + 6 Fe3+ + 7 H2O

Molarity of K2Cr2O7 = number of moles / volume of solution in L

                                                = (0.3525 gm / 294.18 gm mol-1) / 0.250L

                                                = 0.004793 M

Total volume of Fe2+ after dilution with HCl = 100 mL. However, it does not alter the total number of moles of Fe2+ in the solution because Fe2+ does not cause/gain of Fe2+. Therefore, the volume of Fe2+ shall be considered as the original one, 10 mL.

According to the stoichiometry of the reaction, 1 mole K2Cr2O7 neutralizes 6 moles Fe2+. Thus, K2Cr2O7 has 6* oxidizing equivalents.

Now, Using:        M1V1 = M2V2 --- equation 1

Where, M1*= molarity of solution 1, V1= volume of solution 1 (say, K2Cr2O7)

            M2= molarity of solution 2, V2= volume of solution 2 (say, Fe2+)

(6 x 0.004793 M) x 38.40 mL = M2 x 10 mL

Or, 1.104 M mL = M2 x 10 mL

Or, M2 = 1.104 M mL / 10 mL = 0.1104 M

Thus, [Fe2+] = 0.1104 M

Ans. 4. A. 1 mole H2SO4 reacts with 2 mole NH3. So, number of moles of NH3 required is equal to twice the moles of H2SO4.

Using, M1V1 (H2SO4) = M2V2 (NH3)

Or, (2 x 0.250 M) x V1 = 0.350 M x 18 mL

Or, V1 = 6.30 M mL / 0.50 M

Or, V1 = 12.60 mL

Therefore, required volume of 0.250 M H2SO4 = 12.60 mL

Ans. 4B. The number of moles of ammonium sulfate formed is equal to the number of moles of H2SO4 in the specified volume because according to the stoichiometry of the reaction, 1 mol H2SO4 produces 1 mol ammonium sulfate.

Moles of H2SO4 = molarity x volume in L = 0.250 mol L-1 x 0.0126 L = 0.00315 moles

So, moles of ammonium sulfate = 0.00315 moles

Mass of ammonium sulfate = moles x molecular mass = 0.00315 mol x (132.14 g mol-1)

= 0.416 g      

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