Question 35 of 36intermediate🔧 ApplyNumerical5 marks

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O(l) What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Correct Answer

0.9375 g of CaCO₃

Exercise: EXERCISES | Q: 1.35 | (Page 28)
For More Understanding

Explanation

To find the mass of calcium carbonate required, we first determine the number of moles of HCl using its given volume and molarity. Then, using the stoichiometry of the balanced chemical equation, we find the moles of CaCO₃ needed. Finally, we convert the moles of CaCO₃ to mass using its molar mass.

Solution Steps

  1. Step 1: Calculate the moles of HCl Given volume of HCl = 2525 mL = 0.0250.025 L Molarity of HCl = 0.750.75 M Moles of HCl = Molarity ×\times Volume (in L) = 0.75×0.025=0.018750.75 \times 0.025 = 0.01875 mol

  2. Step 2: Relate moles of HCl to moles of CaCO₃ using the balanced equation The reaction is: CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)\text{CaCO}_3 (\text{s}) + 2 \text{HCl} (\text{aq}) \rightarrow \text{CaCl}_2 (\text{aq}) + \text{CO}_2 (\text{g}) + \text{H}_2\text{O}(\text{l}) According to the equation, 11 mole of CaCO3\text{CaCO}_3 reacts with 22 moles of HCl. Therefore, moles of CaCO3\text{CaCO}_3 required = 12×moles of HCl=0.018752=0.009375\frac{1}{2} \times \text{moles of HCl} = \frac{0.01875}{2} = 0.009375 mol

  3. Step 3: Calculate the mass of CaCO₃ Molar mass of CaCO3\text{CaCO}_3 = 40+12+3(16)=10040 + 12 + 3(16) = 100 g/mol Mass of CaCO3\text{CaCO}_3 = moles ×\times molar mass = 0.009375×100=0.93750.009375 \times 100 = 0.9375 g

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