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A small amount of $CaCO_3$ completely neutralized 525ml of N/10 HCl and no acid is left at the end.After converting all calcium chloride to $CaSO_4$,how much plaster of paris can be obtained?

$(a)\;4.2g\qquad(b)\;2.1g\qquad(c)\;3.81g\qquad(d)\;4.38g$

1 Answer

Meq of $CaCO_3$=Meq of HCL=Meq of $CaCl_2$ formed=$525\times \large\frac{1}{10}$$=52.5$
$CaCl_2$ converted to $CaSO_4$ and then to plaster of paris($CaSO_4.\large\frac{1}{2}$$H_2O)$
$\therefore$ Meq of $CaCl_2$ formed=Meq of $CaSO_4$ obtained by $CaCl_2$
$\therefore$ Meq of $CaSO_4.\large\frac{1}{2}$$H_2O$ =Meq of $CaSO_4$ obtained by $CaCl_2$
$\large\frac{w}{\Large\frac{145}{2}}$$\times 1000=52.5$
Weight of $CaSO_4\large\frac{1}{2}$$H_2O=3.81g$
Hence (c) is the correct answer.
answered Jan 7, 2014 by sreemathi.v
 

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