Question

# $$50\ g\ CaCO_{3}$$ are heated to $$1073\ K$$ in a $$5L$$ vessel. What percentage of $$CaCO_{3}$$ would decompose at equilibrium? The value of $$K_{p}$$ for the reaction.$$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$$ is $$1.15\ atm$$ at $$1073\ K$$.

Solution

## Amount of $$CaCO_3 = 50 g$$Molar mass of $$CaCO_3 = 40 + 12 \times 3 \times 16 = 100 gm$$number of mole of $$CaCO_3 = \dfrac{50}{100} = 0.5 mole$$$$CaCO_3 \rightleftharpoons CaO + CO_2 (g)$$$$k_p = P_{CO_2} = 1.15 atm$$volume = $$52$$ temp =  $$1073 k$$$$n = \dfrac{P_{CO_2} v}{RT} = \dfrac{1.15 \times 5 }{0.0821 \times 1073} = 0.065$$$$\%$$ of $$CaCO_3$$ decompose = $$\dfrac{0.065}{5} \times 100$$$$= 1.3\%$$Chemistry

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