CaCO_3 + 2HCl to CaCl_2 + CO_2 + H_2O. In thi | vedclass

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(D) The balanced chemical equation is: $CaCO_3 + 2HCl 	o CaCl_2 + CO_2 + H_2O$Calculate the moles of reactants:$n_{CaCO_3} = \frac{10 \ g}{100 \ g/mo

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$8.325$

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(D) The balanced chemical equation is: $CaCO_3 + 2HCl 	o CaCl_2 + CO_2 + H_2O$Calculate the moles of reactants:$n_{CaCO_3} = \frac{10 \ g}{100 \ g/mol} = 0.1 \ mol$$n_{HCl} = 0.2 \ L 	imes 0.75 \ mol/L = 0.15 \ mol$According to the stoichiometry,$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$. For $0.1 \ mol$ of $CaCO_3$,we need $0.2 \ mol$ of $HCl$. Since we only have $0.15 \ mol$ of $HCl$,$HCl$ is the limiting reagent.Calculate the moles of $CaCl_2$ produced based on the limiting reagent $(HCl)$:$n_{CaCl_2} = \frac{n_{HCl}}{2} = \frac{0.15}{2} = 0.075 \ mol$Calculate the mass of $CaCl_2$:$w_{CaCl_2} = 0.075 \ mol 	imes 111 \ g/mol = 8.325 \ g$

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$. In this chemical reaction,what will be the amount of $CaCl_2$ produced when $10 \ g$ of calcium carbonate and $200 \ mL$ of $0.75 \ M$ $HCl$ are used? $.............. \ g$

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