What mass of 95 % pure CaCO_3 will be require | vedclass

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(A) The balanced chemical equation is:$CaCO_{3(s)} + 2HCl_{(aq)} ightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$Number of moles of $HCl = 	ext{Mol

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$1.32$

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(A) The balanced chemical equation is:$CaCO_{3(s)} + 2HCl_{(aq)} ightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$Number of moles of $HCl = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.5 	imes 0.050 = 0.025 \ 	ext{mol}$.From the stoichiometry of the reaction,$1 \ 	ext{mole}$ of $CaCO_3$ reacts with $2 \ 	ext{moles}$ of $HCl$.Therefore,moles of pure $CaCO_3$ required $= \frac{1}{2} 	imes 0.025 = 0.0125 \ 	ext{mol}$.Mass of pure $CaCO_3 = 	ext{moles} 	imes 	ext{molar mass} = 0.0125 	imes 100 \ 	ext{g/mol} = 1.25 \ 	ext{g}$.Given that the sample is $95 \%$ pure,the mass of the impure sample is calculated as:$	ext{Mass of impure sample} = \frac{	ext{Mass of pure } CaCO_3}{	ext{Purity percentage}} 	imes 100 = \frac{1.25}{95} 	imes 100 \approx 1.3157 \ 	ext{g}$.Rounding to the second decimal place,we get $1.32 \ 	ext{g}$.

What mass of $95 \%$ pure $CaCO_3$ will be required to neutralise $50 \ mL$ of $0.5 \ M \ HCl$ solution according to the following reaction? (In $g$)
$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
[Calculate up to the second decimal place]

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What mass of $95 \%$ pure $CaCO_3$ will be required to neutralise $50 \ mL$ of $0.5 \ M \ HCl$ solution according to the following reaction? (In $g$)$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$[Calculate up to the second decimal place]