How many litres of ethane would be produced when $2.62$ g of vinyl magnesium bromide is treated with $224$ ml of ethyne at STP.

0.224

litre

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

0.08

litre

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

0.448

litre

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

1.12

litre

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

Open in App

Solution

The correct option is C $0.448$ litre
$2 C H_{2} = C H M g B r + C h \equiv C H \to 2 C H_{3} - C H_{3} + M g B r - C \equiv C - M g B r$
milli mole of ethyne = $\frac{224 m l}{22.4 l} = 10 m i l l i m o l$
milli mole of vinyl magnesium bromide $= \frac{2.62}{M} \times 1000$
$= \frac{2.62}{131} \times 1000 = 20 m m o l$
mill mole of ethane generated = $2 \times 10 = 20 m M o l$
$= 20 \times 10^{- 3} \times 22.4$ litre
$= 0.448$ litre

Suggest Corrections

Similar questions

0.595

C H_{3} M g B r

C_{4} H_{9} N H_{2}

0.18 g

A l

d i l . H C l

H_{2}

H_{2}

A l = 27

N H_{3}

4.4 g

C O_{2}

C O_{2}

Join BYJU'S Learning Program