The volume of Cl-2 at STP react with 1-gm-Fe nbsp- to produce FeCl-3 will be1-2-litre1-8-litre0-6-litre2-2-litre

The correct option is C 0.6 litre2Fe(s)+3Cl_2(g)→ 2FeCl_3 (s) moles of Fe=156=0.01786 moles of Cl_2=32× 0.01786=0.02678 Requirement volume of Cl_2 at STP will ...

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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

The volume of...

Question

The volume of $C l_{2}$ at $S T P$ react with $1 gm F e$ to produce $F e C l_{3}$ will be

0.6 litre

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1.2 litre

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1.8 litre

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2.2 litre

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Solution

The correct option is A $0.6 litre$
$2 F e (s) + 3 C l_{2} (g) \to 2 F e C l_{3} (s)$
moles of $F e = \frac{1}{56} = 0.01786$
moles of $C l_{2} = \frac{3}{2} \times 0.01786 = 0.02678$
Requirement volume of $C l_{2}$ at STP will $= 0.02678 \times 22.4 = 0.6 litre$

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The volume of Cl2 at STP react with 1 gm Fe to produce FeCl3 will be

The correct option is A 0.6 litre2Fe(s)+3Cl2(g)→2FeCl3 (s) moles of Fe=156=0.01786 moles of Cl2=32×0.01786=0.02678 Requirement volume of Cl2 at STP will =0.02678×22.4=0.6 litre

The volume of $C l_{2}$ at $S T P$ react with $1 gm F e$ to produce $F e C l_{3}$ will be

The correct option is A $0.6 litre$
$2 F e (s) + 3 C l_{2} (g) \to 2 F e C l_{3} (s)$
moles of $F e = \frac{1}{56} = 0.01786$
moles of $C l_{2} = \frac{3}{2} \times 0.01786 = 0.02678$
Requirement volume of $C l_{2}$ at STP will $= 0.02678 \times 22.4 = 0.6 litre$