A metal oxide has the fomula M_2O_3. It can be reduced by hydrogen to give free metal and water. 0.16g of the metal requires 6mg of H_2for complete reduction. The atomic mass of the metal is .n? times the atomic number of calcium. The value of .n. is

A metal oxide has the fomula M_2O_3. It can be reduced by hydrogen to

1.	A metal oxide has the fomula M_2O_3. It can be reduced by hydrogen to give free metal and water. 0.16g of the metal requires 6mg of H_2for complete reduction. The atomic mass of the metal is .n? times the atomic number of calcium. The value of .n. is
Answer» SOLUTION :`M_(2) O_3 +3H_(2(g)) rarr 2M +3H_2O` 6g gives`H_2`2 moles of metal `:. 6XX10^(-3)` gm of `H_2`GIVE 0.002 moles of .M. `0.002 = (0.16)/("M.Wt")` `:.` M.Wt `=(0.16)/(0.002)=80`