A metal oxide has the formula `A_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. 0.1596 g of this metal oxide requires 6 mg of hydrogen for complete reduction. What is the atomic wight of metal?

Options

A. `52.3`
B. `57.3`
C. `55.8`
D. `59.3`

Answer

C

Explanation

`underset(0.1596 g)(A_(2) O_(3) )+ underset(0.006 g)(3 H_(2)) rarr 2 A + 3H_(2) O`
`0.006 " g of" H_(2)` reduces ` 0. 1596 ` g of ` A_(2) O_(3)`
6 g og `H_(2)` will reduce `(0.1596xx6)/(0.006) = 159.6" g of " A_(2) O_(3)`
Thus, molar mass of `A_(2) O_(3) = 159.6` g
Let, atomic weight of A = x
`therefore 2x + 3 xx 16 = 159.6 `
`2=159.6 - 48 = 111.6`
`x = 55.8`