Calculate how many moles of KMnO4 were present in the volume calculated in b.
b= 25.0 cm3 of FB1 required 35 cm3 of FB2.
FB1 is an aqueous solution containing 21.50 g dm-3 of a mixture of FeSO4 and Fe2(SO4)3.
FB2 is an aqueous solution containing 2.00 dm-3 KMnO4 with a concentration of 0.0127 mol dm-3.

Question

Calculate how many moles of KMnO4 were present in the volume calculated in b.
b= 25.0 cm3 of FB1 required 35 cm3 of FB2.
FB1 is an aqueous solution containing 21.50 g dm-3 of a mixture of FeSO4 and Fe2(SO4)3.
FB2 is an aqueous solution containing 2.00 dm-3 KMnO4 with a concentration of 0.0127 mol dm-3.

anonymous · Answer

Given that 35cm^3 of KMnO4 or FB2 are required for complete reaction, then we use the formula $$n=\frac{ cv }{ 1000 }$$, where n is mol , c concentration, v volume in cm^3.
Therefore, $$n =\frac{ 0.0127*35 }{ 1000 }=0.0004445mol$$