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# How many litres of ethane would be produced when 2.62 g of vinyl magnesium bromide is treated with 224mL of ethyne at STP.A) 0.224LB) 0.08LC) 0.448LD) 1.12L

Last updated date: 11th Jun 2024
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Hint: You first need to write the balanced chemical reaction of vinyl magnesium bromide reacting with ethyne to produce ethene. Then individually calculate moles of each species involved in the reaction. Volume occupied by 1 mole of a substance at STP conditions is 22.4 L.

The required balanced chemical reaction of vinyl lmagnesium bromide with ethyne to produce ethane is:
$2C{H_2} = CH - MgBr + H - C \equiv C - H \to 2C{H_3} - C{H_3} + MgBr - C \equiv C - MgBr$

We are given that the mass of vinyl magnesium bromide ($C{H_2} = CH - MgBr$) is 2.62 g. Volume of ethyne is 224mL.
Molar mass of vinyl magnesium bromide = 131 g
Volume occupied by one mole of any substance at STP conditions is called its molar volume and it is equal to 22.4 L or 22400 mL.

Now, let us calculate moles:
Formula for calculating number of moles = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
Number of moles of vinyl bromide = $\dfrac{{2.62}}{{131}} = 0.02mol$
Number of moles of ethyne = $\dfrac{{{\text{Given volume}}}}{{{\text{Molar volume}}}}$ = $\dfrac{{224}}{{22400}} = 0.01mol$

Now, as you can see in the balanced chemical reaction that 2 moles of vinyl magnesium bromide after reacting with 1 mole of ethyne, producing 2 moles of ethane. Therefore, 0.02 mol of vinyl magnesium bromide will produce 0.02 mol of ethane.
Therefore, moles of ethane produced will be 0.02.
Since, volume of 1 mole = 22.4 L
Volume of 0.02 moles of ethane = 22.4 $\times$ 0.02 = 0.448 L.
So, the correct answer is “Option C”.

Note: Molar volume measured at STP conditions or standard temperature and pressure conditions has a fixed value. At STP conditions, temperature is 273.15 K and pressure is 1 atm or 1 bar. Molar volume is numerically equal to the molar mass divided by mass density.