10 g sample of FeS was roasted and SO 2 thus produced required 10 ml of 0.15 M K 2 Cr 2 O 7 solution in acidic medium. Hence the % of FeS in closest integer in the sample is ............... (Fe =56, S =32 )

Question

10 g sample of FeS was roasted and SO 2 thus produced required 10 ml of 0.15 M K 2 Cr 2 O 7 solution in acidic medium. Hence the % of FeS in closest integer in the sample is ............... (Fe =56, S =32 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

4 FeS +7 O 2( g ) longrightarrow 2 Fe 2 O 3+4 SO 2( g ) Cr 2 O 7 2-+2 H ++3 SO 2( g ) longrightarrow 2 Cr 3++3 SO 4 2-+ H 2 O So moles of K 2 Cr 2 O 7 used =0.15 × 10-2=1.5 × 10-3 mol ∴ moles of SO 2 in the sample =3 × 1.5 × 10-3 mol = mol of FeS ∴ Mass of pure FeS in 10 gram sample =4.5 × 10-3 × 88 g ≈ 4 ∴ % of pure FeS in the sample =3.96 %=4 %

Q. 10g sample of FeS was roasted and SO2​ thus produced required 10ml of 0.15MK2​Cr2​O7​ solution in acidic medium. Hence the % of FeS in closest integer in the sample is ............... (Fe =56,S=32 )

Q. $10 g$ sample of $F e S$ was roasted and $S O_{2}$ thus produced required $10 m l$ of $0.15 M K_{2} C r_{2} O_{7}$ solution in acidic medium. Hence the $%$ of $F e S$ in closest integer in the sample is ............... (Fe $= 56, S = 32$ )