Dilute HCl is reacted with 4.5 moles of calcium carbonate. Calculate : (i) The mass of 4.5 moles of CaCO 3 . (ii) The volume of CO 2 liberated at stp. (iii) The mass of CaCl 2 formed ? (iv) The number of moles of the acid HCl used in the reaction [relative molecular mass of CaCO 3 is 100 and of CaCl 2 is 111.]

CaCO 3 + 2HCl CaCl 2 + H 2 O + CO 2 (i) Given, 1 mole of CaCO 3 = molecular mass of CaCO 3 = 100 g 4.5 moles of CaCO 3 weighs x 4.5 = 450 g (ii) 1 mole of CaCO 3 produces 1 mole of CO 2 and 1 mole occupies 22.4 l of volume. 4.5 moles of CaCO 3 will produce 4.5 moles of CO 2 and 4.5 moles will occupy 22.4 x 4.5 = 100.8 l (iii) 1 mole CaCO 3 produces 111 g. CaCl 2 4.5 moles of CaCO 3 will produce = 111 x 4.5 = 499.5 g. (iv) number of moles of HCl used = 2 x 4.5 = 9 moles. Hence, mass of CaCO 3 = 450 g, vol. of CO 2 liberated = 100.8 lits., mass of Cl 2 formed = 499.5 g, moles of HCl used = 9 moles.

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Chemistry

Dilute HCl is reacted with 4.5 moles of calcium carbonate.
Calculate :
(i) The mass of 4.5 moles of CaCO₃.
(ii) The volume of CO₂ liberated at stp.
(iii) The mass of CaCl₂ formed ?
(iv) The number of moles of the acid HCl used in the reaction [relative molecular mass of CaCO₃ is 100 and of CaCl₂ is 111.]

Stoichiometry

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