calculate the weigh of residue obtained when CuCO3 is strongly heated and 5.6 lit of Co2 is produced at N.T.P
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cuco3------>CUO + CO2
MOLES OF CO2 =5.6/22.4=0.25
1 MOLE OF CO2 CONTAINS 1 MOLE CUCO3
SO
.25 MOLE CO2 CONTAIN .25 MOLE CUCO3
WEIGHT OF CUCO3= 0.25* MOLECULAR WEIGHT
0.25*123.5=30.88GM
MOLES OF CO2 =5.6/22.4=0.25
1 MOLE OF CO2 CONTAINS 1 MOLE CUCO3
SO
.25 MOLE CO2 CONTAIN .25 MOLE CUCO3
WEIGHT OF CUCO3= 0.25* MOLECULAR WEIGHT
0.25*123.5=30.88GM
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Answer:
Explanation:
CuCo3----->CO2+CuO
No. Of moles present in Co2=5.6l/22.4l=0.25mol
1 mole of CaCO3 is required to make 1 mole of CO2
=>0.25 mole of CaCO3 is required to make 0.25 mole of CO2
Molecular mass of CaCO3=40+12+(16×3) =100g
Mass of CaCO3=100×0.25=25g
Mass of CO2=44×0.25=11g
Mass of residue=25g-11g=14g
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