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The mass of 70% H2SO4 required for neutralization of 1 mole of NAOH is????

The mass of 70% H2SO4 required for neutralization of 1 mole of NAOH is????

Grade:12

5 Answers

Dhruvesh Khandelwal
30 Points
9 years ago
The 70% is weight by weight h2so4.
According to the neutralisation reaction
h2so4 + 2NaOH ---> Na2SO4 + 2H2O
the moles of h2so4 needed are 0.5 moles
Molar mass of h2so4 =98g
for 0.5 moles mass of h2so4 = 98*0.5=49g
70 g h2so4 is there in 100 g solution (70% h2so4)
therefore 49g is there in 70g of solution
(49*100)/70=70g
Ans = 70 g
 
nakul
20 Points
9 years ago
but answer is 35 g.....
 
ARNIT
19 Points
7 years ago
0.05 mole of LiAIH, in ether solution was placed in a flask containing 74g (1 mole) of t-butyl alcohol. The product LiAIHCl2H2703 weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li = 7, Al = 27, H = 1,0= 12,0= 16).
Gaurang gupta
21 Points
7 years ago
ApplyPOAC in Li atom....[x/38)=(12.7/254massofcompond)*1 ,x=1.9gms..nw for percentage yield 0.05 mole LiALH4 means 1.9 GMs their for 100%yield
ankit singh
askIITians Faculty 614 Points
3 years ago
= 70 g. required is 70 grams. By Stoichiometry of the reaction: 2 moles of sodium hydroxide is neutralized by 1 mole of sulfuric acid

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