For the following reaction between Mohr\'s salt (iron as FeSO4(NH4)2SO4 6H2O) an

Question

For the following reaction between Mohr's salt (iron as FeSO4(NH4)2SO4 6H2O) and potassium dichromate (dichromate as K2Cr20T), determine the volume (in m ters) of a 0.180 M solution of Mohr's salt that is needed to fully react with 0.0200 L of 0.180 M potassium dichromate. (The reaction is shown in its ionic form in the presence of a strong acid.) Number ml, Mohr's salt volume For the same reaction, what volume (in milliliters) of 0.180 M potassium dichromate is required to fully react with 0.0200 L of a 0.180 M solution of Mohr's salt? Number mL potassium dichromate volume E L

Explanation / Answer

Cr2O7^2- + 6 Fe2+ + 14 H+   ------------> 2 Cr3+ + 6 Fe3+ + 7 H2O

Moles of Cr2O7^2- = Molarity * Volume = 0.180 mol / L * 0.0200 L = 0.0036 mol

Now,

According to the reaction

1 mol of Cr2O7^2- reacts with 6 mol of Fe2+

0.0036 mol of Cr2O7^2- reacts with ( 6 * 0.0036) = 0.0216 mol of Fe2+

Now,

0.0216 mol = 0.180 mol/L * V

V = 0.12 L = 120 mL

Volume of Mohr's salt required = 120 mL

For 2nd case

mol of Fe2+ = 0.0200 L * 0.180 = 0.0036 mol

1 mol of Fe2+ reacts with 1/6 mol of Cr2O7^2-

0.0036 mol of Cr2O7^2- reacts with ( 1/6 * 0.0036) = 0.0006 mol of Cr2O7^2-

Now,

0.0006 mol = 0.180 mol/L * V

V = 0.0033 L = 3.33 mL

Volume of potassium dichromate required = 3.33 mL

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.