20.9285 g sample impure MgCO3 MgO CO2 solid residue 16.7343 g

20.9285 g sample impure MgCO3 MgO CO2 solid residue 16.7343 g

 

 

A 20.9285 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 16.7343 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 = 20.9285 – 16.7343 = 4.1942 g

 

4.1942 g / (44.01 g/mol) = 0.095301 mol CO2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

mass MgCO3

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

( 참고 https://ywpop.tistory.com/19320 )

 

= 4.1942 × (84.31 / 44.01)

= 8.0348 g

 

 

 

답: 8.0348 g