1.

Which of the following statement is/are correct

Answer»

The weight of `CaCO_(3)` in the ORIGINAL MIXTURE is `0.5 g`.
The weight of calcium in the original mixture is `0.2 g`
The weight percent of calcium in the original mixture is 40% `Ca`
The weight percent of `Ca` in the original mixture is 20% `Ca`

SOLUTION :a. Weight of `CaCO_(3) = (0.22 g CO_(2))`
`((1 "MOL" CO_(2))/(44 g CO_(2)))((1 "mol" CaCO_(3))/("mol"CO_(2)))`
`((100 g CaCO_(3))/("mol" CaCO_(3)))`
`= (0.22 xx 100)/(44) = 0.5 g CaCO_(3)`
b. Moles of `CaCO_(3)=` moles of `Ca`
`= ((0.22)/(44)) = 0.005 "mol"`
Weight of `Ca = 0.005 xx 40 = 0.2 g Ca`
d. % of `Ca = (0.2)/(1.0) xx 100 = 20% Ca` ltbgt Hence, (c ) wrong.


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