Question

85g sample of the hydrated sr(oh)2.Xh2o is dried in an oven to give 3.13g of anhydrous sr(oh)2 what is the value of x

Answer 1

Answer: 8

Explanation:

$Sr(OH)_2.xH_2O\rightarrow Sr(OH)_2+xH_2O$

Molar mass of $Sr(OH)_2$ = 122 g/mol

According to stoichiometry:

(122+18x) g of $Sr(OH)_2.xH_2O$ decomposes to give 122 g of $Sr(OH)_2$

Thus 6.85 g of  of $Sr(OH)_2.xH_2O$ decomposes to give =$\frac{122}{122+18x}\times 85g$ of $Sr(OH)_2$

As given :

$\frac{122}{122+18x}\times 6.85g=3.13 g$

[/tex]x=8[/tex]

Thus the value of x is 8.

Answer 2

Answer:

8

Explanation:

below i am giving you the approach to solve the questions . kindly follow that

1- Calculate the mass and moles of H2O driven off:

6.85 g - 3.13 g = 3.72 g H2O / 18.0 g/mol = 0.207 moles H2O

2- Calculate moles Sr(OH)2 remaining:

3.13 g Sr(OH)2 / 121.6 g/mol = 0.0257 moles Sr(OH)2

3- Calculate the ratio of moles H2O / moles Sr(OH)2:

0.207 / 0.0257 = 8.0

so, here the value of x will turns out to be 8 .

HOPE IT CLEARS NOW ..

ALL THE BEST ..