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HETAV PATEL

Grade 11,

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on being heated in oxygen 5.72g of red matallic oxide A was converted to 6.36g black mettalic oxide B.when 4.77g of B was heated in a steam of H2 gas 3.81g of metal M was formed.(given atomic weight of metal-63.5) 1.the formulae of red mettalic oxide A ? 2 the formulae of black metallic oxide? 3 the equivalent weight of metal M in B is?

on being heated in oxygen 5.72g of red matallic oxide A was converted to 6.36g black mettalic oxide B.when 4.77g of B was heated in a steam of H2 gas 3.81g of metal M was formed.(given atomic weight of metal-63.5)
1.the formulae of red mettalic oxide A ?
2 the formulae of black metallic oxide?
3 the equivalent weight of metal M in B is?

Grade:12

1 Answers

Vikas TU
14149 Points
3 years ago
According to the question 
   
  wt of metal obtained on heating 4.77gmof oxide B=3.81gm

wt of oxygen lost=wt of oxide --wt of metal 
                             =4.77-3.81
                             =0.96gm
Equivalent  mass of metal in oxide B=(wt of metal /wt of metal )*8

                                                          =3.81*8/0.96
                                                          =31.50

now valency of metal in B=atomic mass of metal/equivalent wt
                                         =63.50/31.5=2
thus formula of black metal oxide = M+2+O-2=MO

Now since according to the first Question .......

6.36gm Of black oxide obtained by 5.72gm of A(Red Metallic oxide )

4.77gm of B is obtained by (5.72/6.36)*4.77gm A =4.29gm A 

Now we can say that,

3.81gm M obtained by 4.29gm A 

wt of oxygen lost form A = wt of A --wt of metal 

                                     =4.29-3.81=0.48gm

Equivalent wt of metal in A =(3.81/0.48)*8=63.50
Valency of  metal A= Atomic wt /  Equivalent wt 
                                =63.50/63.50=1
Formula of oxide A =M+ + O-2=M2​O

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