CaCO_3 + 2HCl to CaCl_2 + CO_2 + H_2O ; In th | vedclass

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(B) $1$. Calculate the moles of reactants:$Moles \ of \ CaCO_3 = \frac{10 \ g}{100 \ g/mol} = 0.1 \ mol$$Moles \ of \ HCl = Molarity 	imes Volume(L)

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$5.55$

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(B) $1$. Calculate the moles of reactants:$Moles \ of \ CaCO_3 = \frac{10 \ g}{100 \ g/mol} = 0.1 \ mol$$Moles \ of \ HCl = Molarity 	imes Volume(L) = 0.25 \ M 	imes 0.4 \ L = 0.1 \ mol$$2$. Determine the limiting reagent:According to the reaction $CaCO_3 + 2HCl 	o CaCl_2 + CO_2 + H_2O$,$1 \ mol$ 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 $0.1 \ mol$ of $HCl$,$HCl$ is the limiting reagent.$3$. Calculate the product formed:$2 \ mol$ of $HCl$ produces $1 \ mol$ of $CaCl_2$.So,$0.1 \ mol$ of $HCl$ will produce $\frac{0.1}{2} = 0.05 \ mol$ of $CaCl_2$.$4$. Convert moles to mass:$Molar \ mass \ of \ CaCl_2 = 40 + 2 	imes 35.5 = 111 \ g/mol$.$Mass \ of \ CaCl_2 = 0.05 \ mol 	imes 111 \ g/mol = 5.55 \ g$.

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$ ; In this chemical reaction,what will be the amount of calcium chloride formed when $10 \ g$ of calcium carbonate and $400 \ mL$ of $0.25 \ M$ $HCl$ are used? (Atomic masses: $Ca=40, C=12, O=16, H=1, Cl=35.5$)

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