Answer to Question #123486 in General Chemistry for unknown

Question

I react 5g of strontium oxide with 45cm3 of 2M HCl. However, I only produced 1.4g of my salt as somehow my reagents escaped. What is my percentage yield?

Answer 1

Answer

1000 ml 1 (M) HCl = 36.5 gm

45 ml 2 (M) HCl = 3.285 gm

Mole number of 3.285 gm HCl  = 3.285 gm/36.5 gm/mole = 0.09 mole

Molecular weight of Strontium oxide = 103.62 gm

5 gm of strontium oxide = 5 gm/103.62 gm/mole = 0.045 mole

The balanced equation of strontium oxide with HCl is

SrO + 2HCl = SrCl₂ + H₂O

1 mole of strontium oxide will react with 2 mole of HCl to produce 1 mole of SrCl₂.

0.045 mole of strontium oxide will react with (2 x 0.045 = 0.09) mole of HCl to produce 0.045 mole of SrCl₂.

Molecular weight of SrCl₂ is = 153.62 gm

0.045 mole of SrCl₂ is  = (153.62 x 0.045) gm = 7.13 gm

Actual experimental yield is 1.4 gm

Percent yield is (1.4 x 100/7.13)% = 19.63% (Ans.)