Question

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What be the percentage purity of magnsesium carbonate in the sample?
A. 75
B. 96
C. 60
D. 84

Answer 1

Correct Answer - D
Key Concept In the given problem we have provided practical yield of MgO . For calculation of perpentage yield of MgO , we need theoretical yield of MgO . For this we shall use mole concept .
`MgCO_(3)(s)toMgO(s)+CO_(2)(g)` . . . (i)
Moles of `MgCO_(3)=("Weight in gram")/("Molecular weight")= (20)/(84)=0.238 "mol"`
From Eq . (i)
1 mole of `MgCO_(3)` gives =1 mol MgO
`:. 0.238 "mole"MgCO_(3)` will give =0.238 mol MgO
`=0.238xx40g`
`=9.52 g MgO`
Now , practical yield of MgO =8 g
`:.%"Purity"=(8)/(9.52)xx100=84%`
Alternate Method
`underset(84 g)MgCO_(3)tounderset(40 g)MgO+CO_(2)`
`:. 8 g MgO`will be form `(84)/(5) g`
`:. %"Purity"=(84)/(5)xx(100)/(20)=84%`