5.25 g MgCO3 2.37 g MgO percent yield

5.25 g MgCO3 2.37 g MgO percent yield

 

 

A chemist carried out a reaction,

beginning with 5.25 g of magnesium carbonate.

After the reaction was complete,

she obtained 2.37 g of magnesium oxide and carbon dioxide.

a. What was the theoretical yield of MgO in this reaction?

b. What was the percent yield?

 

---------------------------------------------------

 

MgCO3의 몰질량 = 84.31 g/mol 이므로,

5.25 g / (84.31 g/mol) = 0.06227 mol MgCO3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

MgCO3(s) → MgO(s) + CO2(g)

 

MgCO3 : MgO = 1 : 1 계수비(= 몰수비) 이므로,

0.06227 mol MgCO3 반응 시,

생성되는 MgO의 몰수 = 0.06227 mol

 

 

 

MgO의 몰질량 = 40.30 g/mol 이므로,

0.06227 mol × (40.30 g/mol) = 2.51 g MgO

---> 이론적 수득량

 

 

 

(2.37 g / 2.51 g) × 100 = 94.4%

---> 수득률

( 참고 https://ywpop.tistory.com/61 )