The volume of $H_{2}$ needed at $S T P$ for complete hydrogenation of $1.3 g$ of acetylene would be $C_{2} H_{2} + 2 H_{2} \to C_{1} H_{6}$

4.48 d m^{3}

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2.24 d m^{3}

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1.12 d m^{3}

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5.6 d m^{3}

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Solution

The correct option is A $2.24 d m^{3}$
$C_{2} H_{2} + 2 H_{2} \to C_{2} H_{6}$
Amount of acetylene $= 1.3 g$
Molar mass of $C_{2} H_{2} + 2 H_{2} = 12 \times 2 + 2 = 26 g$
Numberof mole of ethylene $= \frac{1.3}{26} = 0.05$

Volume of hydrogen required $= 2 \times 0.05 \times 22.4 l i t r e$
$= 2.24 d m^{3}$

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The volume of H2 needed at STP for complete hydrogenation of 1.3g of acetylene would be C2H2+2H2→C1H6

The correct option is A 2.24dm3C2H2+2H2→C2H6Amount of acetylene =1.3gMolar mass of C2H2+2H2=12×2+2=26g Numberof mole of ethylene=1.326=0.05 Volume of hydrogen required =2×0.05×22.4litre =2.24dm3

The volume of $H_{2}$ needed at $S T P$ for complete hydrogenation of $1.3 g$ of acetylene would be $C_{2} H_{2} + 2 H_{2} \to C_{1} H_{6}$