CaCO3(s) + 2HCl(aq) - CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2 will be formed if 250 mL of 0.76 M HCl reacts with…

🧠 The acid is small, the rock is huge You have $1\,\text{kg}$ of $\ce{CaCO3}$ but only a thin acid solution. The acid will run out long before the rock is touched. So HCl is the limiting reagent, and the salt yield is set by HCl moles.

🗺️ Step by step Moles of HCl $= 0.250 \times 0.76 = 0.19\,\text{mol}$. Moles of $\ce{CaCO3}$ $= 1000/100 = 10\,\text{mol}$.

For $0.19\,\text{mol}$ HCl you need only $0.095\,\text{mol}$ rock — far less than $10$. So HCl limits.

From $\ce{CaCO3 + 2HCl -> CaCl2 + ...}$, $2$ mol HCl gives $1$ mol $\ce{CaCl2}$: $\text{Moles }\ce{CaCl2} = 0.19/2 = 0.095\,\text{mol}.$

Mass of $\ce{CaCl2}$ ($M = 40 + 71 = 111$): $0.095 \times 111 = 10.545\,\text{g}.$

⚠️ Common mistake Don't use rock moles to calculate the yield. The acid is gone first, so it sets the limit.

$\boxed{\text{Answer: (c)}}$