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(A) The balanced chemical equation is:$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_{2}O_{(l)}$Step $1$: Calculate the moles of

Vedclass · Accepted Answer

$0.964$

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(A) The balanced chemical equation is:$CaCO_{3(s)} + 2HCl_{(aq)} ightarrow CaCl_{2(aq)} + CO_{2(g)} + H_{2}O_{(l)}$Step $1$: Calculate the moles of $HCl$ present in $25 \ mL$ of $0.75 \ M$ solution.$Moles = Molarity 	imes Volume (in \ L) = 0.75 \ mol \ L^{-1} 	imes 0.025 \ L = 0.01875 \ mol$ of $HCl$.Step $2$: Use stoichiometry to find the moles of $CaCO_{3}$ required.From the equation,$2 \ mol$ of $HCl$ react with $1 \ mol$ of $CaCO_{3}$.So,$0.01875 \ mol$ of $HCl$ will react with $\frac{0.01875}{2} = 0.009375 \ mol$ of $CaCO_{3}$.Step $3$: Convert moles of $CaCO_{3}$ to mass.Molar mass of $CaCO_{3} = 40 + 12 + (3 	imes 16) = 100 \ g \ mol^{-1}$.$Mass = Moles 	imes Molar \ mass = 0.009375 \ mol 	imes 100 \ g \ mol^{-1} = 0.9375 \ g$ (approximately $0.94 \ g$ or $0.96 \ g$ depending on rounding).Given the options,$0.964 \ g$ is the closest calculated value.

Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_{2}$ and $CO_{2}$ according to the reaction,
$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_{2}O_{(l)}$
What mass of $CaCO_{3}$ is required to react completely with $25 \ mL$ of $0.75 \ M$ $HCl$ (in $g$)?

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Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_{2}$ and $CO_{2}$ according to the reaction,$CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_{2}O_{(l)}$What mass of $CaCO_{3}$ is required to react completely with $25 \ mL$ of $0.75 \ M$ $HCl$ (in $g$)?