Find the per cent composition of each element in below compounds
i. Ca (OH)
ii. MgCO3
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Percentage composition of Ca(OH)2:
The molecular mass of Calcium is 40.1 g/mol.
The molar mass of calcium hydroxide =74g/mol.
1 mole of Ca(OH)2= 74 g Ca(OH)2=40.1 g Ca
The percentage composition of Ca in Ca(OH)2 =40.1×100/74=54%
Percentage composition of MgCO3:
Reaction-
MgCO3____>MgO+CO2
Given that 1 kg MgCO3 which is 90% pure.
Thus actual mass of MgCO3 is 90% of 1 kilogram
=900g
Molecular masses
Magnesium carbonate=24×1+12×1+16×3=84
Magnesium oxide=24×1+16×1=40
Carbon dioxide=12×1+16×2=44
Thus 84 units (grams) of MgCO3 provides 44 units (grams) of CO2
Thus 900 units of MgCO3 will give
(90084×44) units of CO2
=nearly 471.43 units of CO2
Thus,
1kg, 90% pure MgCO3 will give nearly 471.43 grams of CO2
Hope it helped...
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