Question

How many litres of C2H4 could be produced at STP when 2.62 gm of vinyl magnesium bromide is
treated with 224ml of acetylene gas at STP? (Br=80, Mg=24)
(A)
0.224 lit
(B)
0.448 lit
(C) 0.672 lit
(D)
0.112 lit

Answer 1

Given:

• The mass of vinyl magnesium bromide (w) = 2.62 g
• The volume of acetylene gas at STP= 224 mL
• Formula of magnesium vinyl bromide is C₂H₃MgBr
• Molar mass of vinyl magnesium bromide (Mm) = 131 g/mol
• One mole of any gas occupies 22400 mL of volume at STP

To find:

The volume of C₂H₄ in litres that can be produced at STP.

Solution:

• Each acetylene molecule contains two donatable proton ions.
• Two moles of vinyl magnesium bromide reacts with one mole of acetylene to produce two moles of ethene gas.
• Moles of vinyl magnesium bromide = w/Mm = 0.02
• Moles of acetylene = 224/22400 = 0.01
• Moles of ethene gas produced = 0.02
• Volume of ethene gas at STP = 0.02*22.4 = 0.448L

Answer:

The volume of C₂H₄ in litres that can be produced at STP = 0.448 L