Question

an element x reacts with the compound yo3 to produce x3o4. the number of moles x304 produce is 1 mole each of x and yo3 are allowed to react is:
1)1/3
2)2/3
3)1
4)3​

Answer 1

Answer:

Weight of 0.2 mol of X

2

Y

3

=32.0g.

∴ Molecular weight =

0.2

32

=160 g/mol.

Weight of 0.4 mol of X

3

Y

4

=92.8g.

∴ Molecular weight =

0.4

92.8

=232g/mol.

Now suppose the atomic weight of X=a and atomic weight of Y=b

Then 2a+3b=160 and 3a+4b=232.

Solving the equations we get, b=16 and a=56.

Answer 2

Given:

$9 \mathrm{~X}+8 \mathrm{YO}_{3} \longrightarrow 3 \mathrm{~X}_{3} \mathrm{O}_{4}+4 \mathrm{~Y}_{2} \mathrm{O}_{3}$

$9 \text { mole } \mathrm{X} \equiv 8 \text { mole } \mathrm{YO}_{3}$

$\frac{1 \times 9}{8} \text { mole } \mathrm{X} \equiv 1 \text { mole } \mathrm{YO}_{3}$

∴ 1.125 mole X required

but Given 1 mole    'X'  

∴ X is limiting reagent

$9 \text { mole } \mathrm{X} \longrightarrow 3 \text { mole } \mathrm{X}+3 \mathrm{O}_{4}$

$\therefore 1 \text { mole } \mathrm{X} \longrightarrow \frac{3 \times 1}{9}$

$=\frac{1}{3} \text { mole } \mathrm{X}_{3} \mathrm{O}_{4}$

The correct option is "1"

Hence the answer is $\frac{1}{3} \text { mole } \mathrm{X}_{3} \mathrm{O}_{4}$