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 K2Cr207), determine the volume (in mlliiters) of a 0.240 M solution of Mohr's salt that is needed to fully react with 0.0800 L of 0.240 M potassium dichromate. (The reaction is shown in its ionic form in the presence of a strong acid.) 3+ Cr,O 6Fe+14H Number Mohr's salt volume = mL For the same reaction, what volume (in milliliters) of 0.240 M potassium dichromate is required to fully react with 0.0800 L of a 0.240 M solution of Mohr's salt? Number potassium dichromate volume = mL

Explanation / Answer

Cr2O72- + 6 Fe2+ + 14 H+ ---------------> 2 Cr3+ + 6 Fe3+ + 7 H2O

1)

Moles of Cr2O72- = moles of K2Cr2O7

                             = 0.0800 x 0.240

                             = 0.0192 mol

Moles of Mohr's salt = moles of Fe2+ = 6 x moles of (Cr2O7)2-

                               = 6 x 0.0192 = 0.1152 mol

Volume of Mohr's salt = 0.1152 / 0.240

                                     = 0.48 L

Volume of Mohr's salt = 480 mL

(2)

Moles of Fe2+ = moles of Mohr's salt

                        = 0.0800 x 0.240

                       = 0.0192 mol

Moles of K2Cr2O7 = moles of (Cr2O7)2- = 1/6 x0.0192

                              =0.0032 mol

Volume of K2Cr2O7 = 0.0032 /0.240

                               = 0.0133 L

Volume of K2Cr2O7 = 13.3 mL

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