Calculate the number of moles of calcium chloride, \( \mathrm{CaCl}_{2} \), that can be obtained from 25 g of limestone, \( \mathrm{CaCO}_{3} \), in the presence of excess hydrogen chloride, \( \mathrm{HCl} .(\mathrm{Ca}=40, \mathrm{C}=12, \mathrm{O}= \) \( 16, \mathrm{H}=1, \mathrm{Cl}=35.5) \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Bonus Knowledge
To determine the number of moles of calcium chloride, \( \mathrm{CaCl}_{2} \), produced from 25 g of limestone \( \mathrm{CaCO}_{3} \), we begin by calculating the molar mass of \( \mathrm{CaCO}_{3} \). - The molar mass of \( \mathrm{CaCO}_{3} \) = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol. Next, we find the number of moles of \( \mathrm{CaCO}_{3} \) in 25 g: \[ \text{moles of } \mathrm{CaCO}_{3} = \frac{25 \, \text{g}}{100 \, \text{g/mol}} = 0.25 \, \text{mol} \] One mole of \( \mathrm{CaCO}_{3} \) produces one mole of \( \mathrm{CaCl}_{2} \) when reacted with excess \( \mathrm{HCl} \). Therefore, the moles of \( \mathrm{CaCl}_{2} \) produced is also 0.25 mol. In conclusion, from 25 g of limestone, you can obtain **0.25 moles** of calcium chloride, \( \mathrm{CaCl}_{2} \). --- Did you know that limestone, the source of calcium carbonate, has been used for thousands of years in construction and even as a staple in ancient Roman concrete? The Romans relied heavily on limestone for their iconic buildings, many of which still stand today! --- Calcium chloride, a product of this reaction, has many practical uses in the real world. It is commonly used as a de-icing agent on roads in the winter, helping to keep our streets safe, and it also acts as a moisture absorbent in various industries, preventing clumping in products like table salt!
