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(0.1596g)(A_(2)O_(3))+underset(0.06g)(3H_(2))to 2A+3H_(2)O`
0.006 g of `H_(2)` reduces 0.01596 g of ltBrgt `A_(2)O_(3).6g` of `H_(2)` will reduce
`=(0.1596xx6)/(0.006)=159.6g` of `A_(2)O_(3)`
thus, molar mass of `A_(2)O_(3)=159.6g`
Let, atomic weight of A=x
`therefore2x+3xx16=159.6`
`2=159.6-48=111.6`
`x=55.8`