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0.5 mol CaCO3 solid decompose in 500 mL ...

0.5 mol `CaCO_3` solid decompose in 500 mL heated in closed vessel at 400 K reaction `CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))` equilibrium constant of `K_c = 0.9 "mol L"^(-1)`. Calculate mol of `CO_2` at equilibrium how much percentage of reaction completed ?

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`[CO_2]= 0.45 "mol L"^(-1)`, 90% decomposition
`[CO_2]`= 0.3 mol and 60% reaction if `CaCO_3` and CaO used in calculation
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