1.84 g of mixture of CaCO3 and MgCo3 is strongly heated tills further loss in mass takes places. Residue found to weigh 0.96g. Calculate % of each component of mixture.
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Answer:
TRICK IS MORE WEIGHTING COMPOUND IS TAKEN AS X IN REACTANT AND LESS WEIGHTING COMPOUND IS TAKEN Y IN PRODUCT
Explanation:
WHN CACO3 AND MGCO3 ARE HEATED THEY FORM OXIDES AND RELEASE CO2
SO CACO3+MGCO3=MGO+CAO+2CO2 IS EVOLVED
AND LET WEIGHT OF CACO3=X
WEIGHT OF MGCO3 = 1.84-X
WEIGHT OF CAO=0.96-Y
WEIGHT OF MGO = Y
APPLYING POAC FOR CA
MOLES OF CACO3 = MOLES OF CAO
X/100=0.96-Y/56
56X=96-100Y
56X+100Y=96
APPLYING POAC FOR MG
MOLES OF MGCO3 = MOLES OF MGO
1.84-X /84 = Y/40
73.6-40X=84Y
40X+84Y=73.6
ON SOLVING Y=0.4
AND X = 1
WEIGHT OF CACO3 = 1
WEIGHT OF MGCO3 = 0.84
PERCENTAGE COMPOSITION OF CACO3 = 100/1.84=54.34
PERCENTAGE COMPOSITION OF MGCO3 = 84/1.84=45.65
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