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what weight of 70% pure sulphuric acid is require for complete neutralisation of 4 gram of sodium hydroxide?

what weight of 70% pure sulphuric acid is require for complete neutralisation of 4 gram of sodium hydroxide?
 

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
Lets consider the balance response 
H2SO4 + 2NaOH  Na2SO4 + 2H2O 
From this condition the moles of H2SO4 required here is 0.5 moles 
Molar mass of H2SO4 = 2* H + S +4 *O 
= 2*1 + 32 + 4*16 
= 98g 
Gram nuclear mass = Molar mass * moles 
= 98 * 0.5 
= 49 g 
As in the given inquiry, weight of 70% immaculate sulphuric corrosive is require for finish balance of 4 gram of sodium hydroxide 
Here 49 g is accessible in 70g of the arrangement 
At that point (49*100)/70 = 70g 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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