An equimolar mixture of ferrous oxalate & stannous chloride is treated with decinormal acidic KMnO4 giving ferric, stannic & chlorate ions along with CO2 gas. If the titre value is 225 ml. Numbers of millimoles of chlorate ions produced are

Let millimoles of FeC2 O4 =SnCl2 = xn -factor of FeC2 O4 =3n - factor of SnCl2 = 14 ∴3x+14x=0.1×225⇒17x=22.5⇒x=22.517m.moles of ClO3−formed =22.517×2=2.64

An equimolar mixture of ferrous oxalate & stannous chloride is treated with decinormal acidic KMnO4 giving ferric, stannic & chlorate ions along with CO2 gas. If the titre value is 225 ml. Numbers of millimoles of chlorate ions produced are

Let millimoles of FeC2 O4 =SnCl2 = xn -factor of FeC2 O4 =3n - factor of SnCl2 = 14 ∴3x+14x=0.1×225⇒17x=22.5⇒x=22.517m.moles of ClO3−formed =22.517×2=2.64

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

An equimolar mixture of ferrous oxalate & stannous chloride is treated with decinormal acidic KMnO₄giving ferric, stannic & chlorate ions along with CO₂ gas. If the titre value is 225 ml. Numbers of millimoles of chlorate ions produced are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2.64 m moles

1.5 m moles

3 m moles

7.5 m moles

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let millimoles of FeC₂ O₄ =SnCl₂ = x
n -factor of FeC₂ O₄ =3
n - factor of SnCl₂ = 14 $∴ 3 x + 14 x = 0.1 \times 225$
$\Rightarrow 17 x = 22 .5 \Rightarrow x = \frac{22 .5}{17}$
m.moles of ${ClO}_{3}^{-} formed = \frac{22.5}{17} \times 2 = 2.64$