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The mass of 70% Sulphuric acid by mass required to neutralize 1 mole of NaOH

Asked by 21st March 2013, 10:35 AM
Answered by Expert
Answer:

2NaOH  + H2SO4 = Na2SO4 + H2O

Pure H2SO4 required for 1 mole of NaOH =  1/2 mole = 49g

70% H2SO4 required for 1 mole = 49 x 100 /70

                                                  = 70 g
Answered by Expert 3rd May 2013, 7:35 PM
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