Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction given below:

     CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

What mass of CaCl₂ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃? Name the limiting reagent. Calculate the number of moles of CaCl₂ formed in the reaction.

<p>This is a Long Answer Type Questions as classified in NCERT Exemplar</p><p>The volume of HCl solution is 250 mL and its molarity is 0.76M.</p><p>The number of moles of HCl as follows, </p><p>Moles of HCl Molarity Volume (in L)</p><p>= 0.76M 0.250 L</p><p>= 0.19 mol</p><p>The molar mass of CaCO₃ is 100 g / gQl and the mass of CaCO₃ is given as 1000 g</p><p>The number of moles of CaCO_{3 i}s calculated as</p><p>Moles of CaCO₃ = &nbsp; &nbsp; M a s&nbsp; / M o l a r &nbsp; m a s &nbsp; &nbsp; &nbsp;&nbsp;</p><p>= &nbsp; &nbsp; &nbsp;1000 g / 100 g &nbsp; / &nbsp; m o l&nbsp; = 10 mol</p><p>According to the given reaction, 1 mole of CaCO₃requires 2 moles of HCl. So, the required number of moles of HCl for 10 moles of CaCO₃ is calculated as</p><p>Moles of HCl =&nbsp; &nbsp;2mol of HCl /1mol of CaCO₂&nbsp; &nbsp;* 10 mol of CaCO₃</p><p>= &nbsp;20 mol</p><p>So, the required number of moles of HCl is 20 mol but only 0.19 mol are given. So, HCl is a limiting reagent. So, the amount of calcium chloride formed is depend upon the limiting reagent, that is, the amount of HCl available.</p><p>According to the reaction, 2 moles of HCl gives 1 mol of CaCl₂ Sg 2 ., the number of moles of calcium chloride produced by 0.19 mol of HCl as follows, </p><p>Moles of CaCl₂ =&nbsp; 1 mol of CaCl₂ / 2 mol&nbsp; of HCl&nbsp; x 0.19 mol of HCl</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 0.095 mol</p><p>The molar mass of calcium chloride is 111 g / mol. So, its mass is calculated as</p><p>Mass of CaCl₂ = Moles x &nbsp;Molar mass</p><p>= 0.095 mol x 111 g / mol</p><p>= 10.54 g</p>

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