The volume of Cl 2 at STP react with 1 gm Fe to produce F eCl 3 will beA- 1-2 litreB- 1-8 litreC- 0-6 litreD- 2-2 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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.2 litre

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.8 litre

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.2 litre

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $= 0.02678 \times 22.4 = 0.6 litre$

Suggest Corrections

Similar questions

0.18 g

A l

reacts with

d i l . H C l

to give

H_{2}

. The volume of

H_{2}

evolved at STP is:
(Atomic weight of

A l = 27

)

Calcium carbonate decomposes on heating according to the equation:

$C a C O_{3}$ (s) $⟶$ CaO(s)+ $C O_{2}$

At STP, the volume of $C O_{2}$ obtained by thermal decomposition of 50g of $C a C O_{3}$ will be:

What will be the volume of $C O_{2}$ at NTP obtained on heating 10 grams of ( $90 %$ pure) limestone

Q. A certain gas occupies 0.418 litre at

27^{\circ} C

and 740mm Hg. What is the Volume at S.T.P.?

Q. A certain gas occupies 0.418 litre at $27^{\circ}C$ and 740mm Hg. What is the Volume at S.T.P.?

Join BYJU'S Learning Program

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 $= 0.02678 \times 22.4 = 0.6 litre$