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Question

0.5 moles of BaCl2 is mixed with 0.2 moles of (NH4)3PO4. Reactant present in excess (with extra number of moles of that reactant) is :

A
BaCl2,0.3mol
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B
(NH4)3PO4, 0.2mol
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C
(NH4)3PO4,0.1mol
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D
BaCl2, 0.2mol
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Solution

The correct option is D BaCl2, 0.2mol
3BaCl2+2(NH4)3PO4Ba3(PO4)2+6NH4Cl
3mol2mol1mol
Given: 0.5mol0.2mol0.1mol
Thus, (NH4)3PO4 is the limiting reactant giving 0.1 mol Ba3(PO4)2.
BaCl2 required by 0.2 mol of (NH4)3PO4=0.3mol.
Thus, BaCl2 in excess =0.50.3=0.2mol

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