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32g of CaC2 react at 27°C with excess of water to produce acetylene gas. What is work done?
a)-2400 Cal b)-300 Cal c)-300 J d)-2400 L-atm
Please explain with detailed explanation
Write down the balanced reaction.
CaC2 + 2 H2O (excess) → C2H2 + Ca(OH)2
Molar Mass of CaC2 = 40 + (12*2) = 40 + 24 = 64 g/mol
No. of Mole of CaC2 = 32/64 = 0.5 mole
Now, Work Done = ΔngRT,
where, Δng is difference in no. of (gaseous) mole of reactant and product;
R is the Universal Gas Constant;
T is the absolute temperature.
Thus, W = (-0.5)(8.314)(300) = 1247.1 J
or W = (-0.5)(0.0821)(300) = 12.315 L-atm
or W = (-0.5)(1.987)(300) = 298.05 cal ≈ 300 cal
Hence option (b) is correct.
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