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InterviewSolution
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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? |
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Answer» (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 (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 |
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