0.42 gram of metal carbonate is neutralized by 50cc ,0.2N of HCL solution. Calculate equivalent weight of the metal
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11th
Chemistry
Some Basic Concepts of Chemistry
Stoichiometry and Stoichiometric Calculations
25 mL of a solution of Na2C...
CHEMISTRY
Asked on December 20, 2019 byMitali Bains
25 mL of a solution of Na2CO3 having a specific gravity of 1.25 g/mL required 32.9 mL of a solution of HCl containing 109.5 g of the acid per litre for complete neutralization. The volume (in mL) of 0.84 N H2SO4 that will be completely neutralized by 125 g of Na2CO3 solution is :
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ANSWER
The specific gravity of sodium carbonate is 1.25 g/mL.
Hence, the mass of 25 mL of solution of sodium carbonate is 1.25×25=31.25 g.
The mass of HCl used for neutralization is 100032.9×109.5=3.60 g.
The number of moles of HCl in 3.60 g is 36.53.60=0.0987.
2 moles of HCl will neutralize 1 mole of sodium carbonate.
The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are 20.987=0.04935
Thus, 31.25 g of sodium carbonate contains 0.04935 moles.
Hence, 125 g of sodium carbonate will contain 31.25125×0.04935=0.1974 moles.
They will neutrlaize 0.1974 moles of sulphuric acid which corresponds to 2×0.1974=0.3948 g eq of sulphuric acid.
The volume of 0.84 N sulphuric acid required will be 0.84
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