Question

A semiconductor YBa2Cu3o7 is prepared by a reaction involving Y2O3, BaO2 and CuO. The ratio of their moles should be.

Answer 1

Given:

A semiconductor YBa₂Cu₃O₇ is prepared from the compounds Y₂O₃, BaO₂ and CuO

To find:

The ratio of moles of Y₂O₃, BaO₂ and CuO

Calculation:

The following reaction is involved in the preparation of inorganic compound

YBa₂Cu₃O₇

  $\frac{1}{2}Y_{2}O_{3}+2BaO_{2}+3CuO\longrightarrow YBa_{2}Cu_{3}O_{7}+\frac{3}{4} O_{2}$

For preparation of 1 mole of YBa₂Cu₃O₇ it requires

1. 0.5 moles of Y₂O₃
2. 2 moles of BaO₂
3. 3 moles of CuO

Therefore the ratio of moles is

Y₂O₃:BaO₂:CuO = 0.5:2:3

                               = 1:4:6

Final answer:

The ratio of moles of Y₂O₃, BaO₂ and CuO is 1:4:6