1. A volume of 25.00 mL of a manganese(II) solution is titrated in basic conditi

Question

1. A volume of 25.00 mL of a manganese(II) solution is titrated in basic conditions with a volume of 34.77 mL of 0.05876 M KMnat. Determine the concentration of manganese(II) ions in the solution. (3.5 pts) Mn2+(aq) + Mn04-(aq) --+ MnO(s) not balanced

2. Determine the enthalpy change for the following reaction:

C2F14(g) + H2(g) -- C2H6(g)

Use appropriate data from the following listing:

C2H4(g) ± 302(g) --, 2CO2(g) ± 21420(1) AH1 = - 1410.9 kJ

2C2H6(g) + 702(g) —› 4CO2(g) + 6H20(1) AH2= - 3119.4 kJ

2H2(g) + 02(g) —> 2H20(g) AH3= -483.6 kJ

2F12(g) + 02(g) —> 2H20(1) A1I4= - 571.6 kJ

H20(1) — H20(g) AH5 7-- + 43.98 kJ

Explanation / Answer

Ans 1 :

The balanced reaction is given as :

Mn2+ + 2MnO4- = 3MnO

Number of moles of KMnO4 = 0.05876 x 0.03477 = 0.00204 moles

So the number of moles of Manganes ions = 2 x 0.00204

= 0.00408 moles

Concentration = no. of moles / volume of solution in L

= 0.00408 / 0.025

= 0.1634 M

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