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`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» Correct Answer - `(i)51.66g" "(ii)37.31g` `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` |
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