21.8603 g sample impure MgCO3 MgO CO2 solid residue 18.7358 g

21.8603 g sample impure MgCO3 MgO CO2 solid residue 18.7358 g

 

 

A 21.8603 g sample of impure magnesium carbonate

was heated to complete decomposition according to the equation

MgCO3(s) → MgO(s) + CO2(g).

After the reaction was complete,

the solid residue (consisting of MgO and the original impurities)

had a mass of 18.7358 g.

Assuming that only the magnesium carbonate had decomposed,

how much magnesium carbonate was present in the original sample?

Answer in units of g.

 

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> MgCO3의 몰질량 = 84.31 g/mol

> CO2의 몰질량 = 44.01 g/mol

 

 

 

CO2의 질량 = 21.8603 – 18.7358 = 3.1245 g

 

3.1245 g / (44.01 g/mol) = 0.070995 mol CO2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

MgCO3(s) → MgO(s) + CO2(g)

 

MgCO3 : CO2 = 1 : 1 계수비(= 몰수비) 이므로,

분해된 MgCO3의 몰수 = 0.070995 mol

 

0.070995 mol × (84.31 g/mol) = 5.9856 g MgCO3

---> 시료에 들어있는 순수한 MgCO3의 질량

 

 

 

 

mass MgCO3

= mass CO2 × (molar mass MgCO3 / molar mass CO2)

= 3.1245 × (84.31 / 44.01)

= 5.9856 g

 

 

 

답: 5.9856 g