Chemical maxima

Chemistry Level 3

In a beaker conatining 50 50 gm of pure chalk, 46 46 gm of sodium and 73 73 gm of pure hydrochloric acid are added. Then 9 9 gm of pure water, 35.5 35.5 gm of chlorine molecule and 4 4 gm of carbon atoms are removed from the beaker. If all these reactants react to form a chemical compund without any wastage of any reactant, then what is the maximum number of moles that can be present in the reacting compound?


The answer is 1.167.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Apr 23, 2016

Gram molecular mass of C a C O 3 CaCO_3 is 40 + 12 + 3 ( 16 ) = 100 40 + 12 + 3(16) = 100 gm
So, number of moles of C a C O 3 CaCO_3 is 50 100 = 1 2 \dfrac{50}{100} = \dfrac{1}{2}

Gram atomic mass of N a Na is 23 23 gm
So, number of moles of N a Na is 46 23 = 2 \dfrac{46}{23} = 2

Gram molecular mass of H C l HCl is 1 + 35.5 = 36.5 1 + 35.5 = 36.5 gm
So, number of moles of H C l HCl is 73 36.5 = 2 \dfrac{73}{36.5} = 2

Gram atomic mass of C C is 12 12 gm
So, number of moles of C C is 4 12 = 1 3 \dfrac{4}{12} = \dfrac{1}{3}

Gram molecular mass of C l 2 {Cl}_2 is 2 × 35.5 = 71 2 × 35.5 = 71 gm
So, number of moles of C a C O 3 CaCO_3 is 35.5 71 = 1 2 \dfrac{35.5}{71} = \dfrac{1}{2}

Gram molecular mass of H 2 O H_2O is 2 ( 1 ) + 16 = 18 2(1) + 16 = 18 gm
So, number of moles of H 2 O H_2O is 9 18 = 1 2 \dfrac{9}{18} =\dfrac{1}{2}

So, the maximum number of moles of the resulting element is ( 1 2 + 2 + 2 ) − ( 1 3 + 1 2 + 1 2 ) (\dfrac12 + 2 + 2) - (\dfrac13 + \dfrac12 + \dfrac12)
= 2 1 2 − 1 1 3 2\dfrac12 - 1\dfrac13
= 7 6 \dfrac76
= 1.167 \boxed{1.167} moles.

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