4.5 moles of calcium carbonate are reacted with dilute hydrochloric ac

1.	4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid. (i) Write the equation for the reaction. (ii) What is the mass of 4.5 moles of calcium carbonate ? (Relative molecular mass of calcium carbonate is 100). (iii) What is the volume of carbon dioxide liberated at stp ? (iv) What mass of calcium chloride is formed ? (Relative molecular mass of calcium chloride is 111). (v) How many moles of HCI are used in this reaction ?
Answer» (i) CaCO₃ + 2HCl ---> CaCl₂ + H₂O + CO₂ (ii) Mass of 1 more of CaCO₃ = 100 g Mass of 4.5 moles of CaCO₃ = 4.5 x 100 = 450 g (iii) 1 mole CaCO₃liberates 1 mole of CO₂ .'. 4.5 moles of CaCO₃ liberates 4.5 moles of CO₂ Volume of CO₂ liberated at STP = 22.4 x 4.5 = 100.8 l. (iv) 1 mole of CaCO₃ gives 1 mole of CaCl₂. .'. 4.5 moles of CaCO₃ gives 4.5 moles of CaCl₂ = 4.5 x 111 = 499.5 g (v) 1 mole of CaCO₃ requires 2 moles of HCl 4.5 moles of CaCO₃ requires = 2 x 4.5 = 9 moles of HCl

4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid. (i) Write the equation for the reaction. (ii) What is the mass of 4.5 moles of calcium carbonate ? (Relative molecular mass of calcium carbonate is 100). (iii) What is the volume of carbon dioxide liberated at stp ? (iv) What mass of calcium chloride is formed ? (Relative molecular mass of calcium chloride is 111). (v) How many moles of HCI are used in this reaction ?

Answer»

(i) CaCO₃ + 2HCl ---> CaCl₂ + H₂O + CO₂

(ii) Mass of 1 more of CaCO₃ = 100 g

Mass of 4.5 moles of CaCO₃ = 4.5 x 100 = 450 g

(iii) 1 mole CaCO₃liberates 1 mole of CO₂

.'. 4.5 moles of CaCO₃ liberates 4.5 moles of CO₂

Volume of CO₂ liberated at STP = 22.4 x 4.5 = 100.8 l.

(iv) 1 mole of CaCO₃ gives 1 mole of CaCl₂.

.'. 4.5 moles of CaCO₃ gives 4.5 moles of CaCl₂ = 4.5 x 111 = 499.5 g

(v) 1 mole of CaCO₃ requires 2 moles of HCl

4.5 moles of CaCO₃ requires = 2 x 4.5 = 9 moles of HCl