Oleum is a mixture of
\(H_{2}SO_{4}\)+\(SO_{2}\)
\(H_{2}SO_{4}\)+\(SO_{3}\)
\(H_{2}S_{2}O_{3}\)+\(SO_{3}\)
\(H_{2}S_{2}O_{3}\)+\(SO_{2}\)

Which of the following is a disproportionation reaction \(4Zn+10HNO_{3}\rightarrow4Zn(NO_{3})_{2}+NH_{4}NO_{3}+3H_{2}O\) \(2KBr+3H_{2}SO_{4}\rightarrow2KHSO_{4}+SO_{2}+Br_{2}+2H_{2}O\)\(2KMnO_{4}+5H_{2}S+4H_{2}SO_{4}\rightarrow2KHSO_{4}+2MnSO_{4}+5S+8H_{2}O\),\(S_{8}+NaOH \rightarrow Na_{2}S_{5}+Na_{2}S_{2}O_{3}+H_{2}\)

What is the O.N in (a)H_(2)S (b) SO_(2) (c ) SO_(3)?

Draw the structures of (i) H_(3)PO_(2) " " (ii) H_(2)SO_(5) " " (iii) H_(2)SO_(3) " " (iv) H_(2)SO_(4) " " (v) H_(2)S_(2)O_(7) .

H_(4)underline(S_(2))O_(6)+H_(2)O to H_(2)SO_(3)+H_(2)SO_(4)

Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentage composition of H_(2)SO_(4), SO_(3) (free) and SO_(3) (combined) is calculated. Oleum is nothing but it is a mixture of H_(2)SO_(4) and SO_(3) i.e., H_(2)S_(2)O_(7) , which is obtained by passing. SO_(3) in solution of H_(2)SO_(4) . In order of dissolve free SO_(3) in oleum, dilution of oleum is done, in which oleum converts into pure H_(2)SO_(4) . It is shown by the reaction as under : H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure") or " " SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure") When 100g sample of oleum is diluted with desired weight of H_(2)O("in" g) , then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as ""109%H_(2)SO_(4)" it means that 100 g of oleum on dilution with 9m of H_(2)O provides 109g pure H_(2)SO_(4) , in which all free SO_(2) in 100g of oleum is dissolved. In the above question number 1 , what is the percentage of free SO_(3) and H_(2)SO_(4) in the oleum simple respectively ?

Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentage composition of H_(2)SO_(4), SO_(3) (free) and SO_(3) (combined) is calculated. Oleum is nothing but it is a mixture of H_(2)SO_(4) and SO_(3) i.e., H_(2)S_(2)O_(7) , which is obtained by passing. SO_(3) in solution of H_(2)SO_(4) . In order of dissolve free SO_(3) in oleum, dilution of oleum is done, in which oleum converts into pure H_(2)SO_(4) . It is shown by the reaction as under : H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure") or " " SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure") When 100g sample of oleum is diluted with desired weight of H_(2)O("in" g) , then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as ""109%H_(2)SO_(4)" it means that 100 g of oleum on dilution with 9m of H_(2)O provides 109g pure H_(2)SO_(4) , in which all free SO_(2) in 100g of oleum is dissolved. For 109% labelled oleum if the number of moles of H_(2)SO_(4) and free SO_(3) be x and y respectively, then what will be the value of (x+y)/(x-y) ?

Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentage composition of H_(2)SO_(4), SO_(3) (free) and SO_(3) (combined) is calculated. Oleum is nothing but it is a mixture of H_(2)SO_(4) and SO_(3) i.e., H_(2)S_(2)O_(7) , which is obtained by passing. SO_(3) in solution of H_(2)SO_(4) . In order of dissolve free SO_(3) in oleum, dilution of oleum is done, in which oleum converts into pure H_(2)SO_(4) . It is shown by the reaction as under : H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure") or " " SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure") When 100g sample of oleum is diluted with desired weight of H_(2)O ("in" g) , then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as ""109%H_(2)SO_(4)" it means that 100 g of oleum on dilution with 9m of H_(2)O provides 109g pure H_(2)SO_(4) , in which all free SO_(2) in 100g of oleum is dissolved. In the above question number 1 , what will be the percentage of combined SO_(3) in the given oleum sample?

Balance the following equations : (i) KMnO_(4) + H_(2)SO_(4) + H_(2)O_(2) to K_(2)SO_(4) + MnSO_(4) + H_(2)O + O_(2) (ii) KMnO_(4) + KCl + H_(2)SO_(4) to MnSO_(4) + K_(2)SO_(4) + H_(2)O + Cl_(2) (iii) MnO_(2) + H_(2)O_(2) to MnO_(4)^(-) + MnO_(4)^(-) + H_(2)O (Basic medium)