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If 6.3 g of NaHCO(3) are added to 15.0 g...

If 6.3 g of NaHC`O_(3)` are added to 15.0 g `CH_(3)`COOH solution, the residue is found of weight 18.0 g. What is the mass of `CO_(2)` released in the reaction?

Text Solution

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`underset(6.3 g)(NaHCO_(3)) + underset(15.0 g)(CH_(3)COOH) to underset("Residue 18.0g)(CH_(3)COONa + H_(2)O) + underset(xg)(CO_(2))`
Sum of the mass of reactants = Mass of `NaHCO_(3)` + Mass of `CH_(3)COOH`
`=6.3 + 15.0 = 21.3`g
Sum of the mass of products = Mass of residue + Mass of `CO_(2)`
`(180.0 + x)g`
(where x is mass of `CO_(2)` released)
According to law of conservation of mass
Mass of reactants = Mass of products
21.3 = 18.0 +x
or x=21.3-18.0 = 3.3 g
`therefore` Mass of `CO_(2)` released = 3.3 g
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