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

Question

For the following reaction between Mohr's salt (iron as FeSO4(NH4)2SO4·6H2O) and potassium dichromate (dichromate as K2Cr2O7), determine the volume (in milliliters) of a 0.130 M solution of Mohr's salt that is needed to fully react with 0.0500 L of 0.130 M potassium dichromate. (The reaction is shown in its ionic form in the presence of a strong acid.)

A) Mohr's salt volume in mL

B)For the same reaction, what volume (in milliliters) of 0.130 M potassium dichromate is required to fully react with 0.0500 L of a 0.130 M solution of Mohr's salt?

Explanation / Answer

A) The reaction between Mohr's salt (iron as FeSO4(NH4)2SO4·6H2O) and potassium dichromate (dichromate as K2Cr2O7) is as follows:

Cr2O2-7 + 6Fe2+ + 14 H+  2Cr3+ +6 Fe3+ +7H2O

Here number of moles of K2Cr2O7=

0.130 M K2Cr2O7 x 0.0500 L = 0.0065 moles K2Cr2O7

0.0065 moles K2Cr2O7 x (6Fe2+ / 1 Cr2O7 2-) = 0.039 moles Fe2+
0.0336moles Fe2+ / 0.13 M Fe2+ = 0.3 L or 300 ml

B)

The reaction between Mohr's salt (iron as FeSO4(NH4)2SO4·6H2O) and potassium dichromate (dichromate as K2Cr2O7) is as follows:

Cr2O2-7 + 6Fe2+ + 14 H+  2Cr3+ +6 Fe3+ +7H2O

Here number of moles of Mohr's salt =

0.130 M K2Cr2O7 x 0.0500 L = 0.0065 moles Mohr's salt

0.0065 moles Mohr's salt 7 x (1 Cr2O7 2-/6Fe2+) = 0.00108 moles Cr2O7 2-/
0.00108 moles Cr2O7 2-/ / 0.13 M Cr2O7 2-/ = 0.0083 L or 0.833 ml

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