Question

4.5 moles of calcium carbonate are reached with dilute hydrochloric acid.

 (i) Write the equations 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 HCl are used in this reaction?

Answer 1

(i) CaCO₃ + 2HCL ⟶ CaCl₂ + H₂O + CO₂

(ii) Relative molecular mass of CaCO₃ = 100 

If mass of 1 mole of CaCO₃ = 100 gm

Then 4.5 moles of CaCO₃ = \(\frac{100}{1}\times 4.5\) = 450 gm.

(iii) Molar volume of a gas at STP = 22.4l

So, Vol. of CO₂ at STP = 22.4l

(iv) Relative molecular mass of CaCl₂ = 111

If mass of 1 mole of CaCl₂ = 111 gm

Then 4.5 moles of CaCl₂ =  \(\frac{111}{1}\times4.5\)  = 499.5 gm.

(v) According to equation 

CaCO₃ + 2HCL ⟶ CaCl₂ + H₂O + CO₂

       1   :     2                1

No. of moles of HCL used = 2

No. of molecules = No. of moles x 6 x 10²³

                           = 2 x 6 x 10²³

                          =  12 x 10²³

Answer 2

(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 CaCO3 liberates 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