How many grams of CaCO_3 must be decomposed t | vedclass

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(C) The decomposition reaction is: $CaCO_3(s) 	o CaO(s) + CO_2(g)$.From the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.Using th

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$80$

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(C) The decomposition reaction is: $CaCO_3(s) 	o CaO(s) + CO_2(g)$.From the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.Using the ideal gas law $PV = nRT$,we calculate the moles of $CO_2$:$n = \frac{PV}{RT} = \frac{(750/760) \ atm 	imes 20 \ L}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 300 \ K} \approx 0.803 \ mol$.Rounding to the nearest significant value,$n \approx 0.8 \ mol$.Mass of $CaCO_3 = n 	imes 	ext{Molar Mass} = 0.8 \ mol 	imes 100 \ g/mol = 80 \ g$.

How many grams of $CaCO_3$ must be decomposed to produce $20 \ L$ of $CO_2$ at $750 \ torr$ and $27 \ ^\circ C$?

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