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  •   The amount of BaSO4 formed upon mixing100 mL of 20.8% BaCl2 solution with    50 mL of 9.8% H2SO4 solution will be :  (B

  The amount of BaSO4 formed upon mixing100 mL of 20.8% BaCl2 solution with   
50 mL of 9.8% H2SO4 solution will be :  (Ba=137, Cl=35.5, S=32, H=1 and O=16)                

 

Option: 1

  23.3    g  
 


Option: 2

 11.65 g   
 


Option: 3

30.6 g  
 


Option: 4

33.2    g 


Answers (1)

best_answer

Moles of \mathrm{BaCl_{2}=\frac{20.8g}{208}=\textup{0.1ml moles}\:of\:H_{2}SO_{4}=\frac{9.8}{98}\times \frac{1}{2}=0.05ml}

\\ \mathrm{BaCl_{2}=\frac{20.8g}{208}=0.1ml }\\ \mathrm{moles\:of\:H_{2}SO_{4}=\frac{9.8}{98}\times \frac{1}{2}=0.05ml}

\\\mathrm{ BaCl_{2}+H_{2}SO_{4}\rightarrow BaSO_{4}+2H^{+}+2Cl^{-} }\\ \\\mathrm{ BaSO_{4}\:formed =0.05ml\\ =0.05\times 11.65g}

Posted by

Deependra Verma

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