`50 g` of `CaCO_(3)` is allowed to react with `70 g` of `H_(3)PO_(4^(.)` Calculate :
`(i)` amount of `Ca_(3)(PO_(4))_(2)` formed
`(ii)` amount of unreacted reagent

Answer

`(i)51.66g" "(ii)37.31g`

Explanation

`3CaCO_(3)+2H_(3)PO_(4)rarrCa_(3)(PO_(4))_(2)+3H_(2)O+3CO_(2)`
`{:(50//100"mole",70//98"mole"),(=0.5,0.7142),(" "-,0.7142-(2)/(3)xx0.5=0.3808((0.5)/(3))):}`
Limiting reactant
`m_(CaCO_(3))=(0.5)/(3)xxM_(Ca_(3)(PO_(4))_(2))=51.66gm`
`m_(H_(3)PO_(4))=0.3808xxM_(H_(3)PO_(4)=31.31gm`