If 0.6 mole of MgCl 2 is mixed with 0.4 mole of K 3 PO 4 , the maximum number of moles of magnesium phosphate, that can be obtained is :- (Atomic masses : Mg = 24 , P = 31 , O = 16 , Cl = 35.5 )

Question

If 0.6 mole of MgCl 2 ​ is mixed with 0.4 mole of K 3 ​ PO 4 ​ , the maximum number of moles of magnesium phosphate, that can be obtained is :- (Atomic masses : Mg = 24 , P = 31 , O = 16 , Cl = 35.5 )

TestHub · Accepted Answer

To determine the maximum moles of magnesium phosphate, first write the balanced chemical equation:

$3 MgCl_{2} + 2 K_{3} PO_{4} \to Mg_{3} (PO_{4})_{2} + 6 KCl$

Calculate moles of product from each reactant:

From $MgCl_{2}$ : $0.6 m o l MgCl_{2} \times \frac{1 m o l Mg _{3} ( PO _{4} ) _{2}}{3 m o l MgCl _{2}} = 0.2 m o l Mg_{3} (PO_{4})_{2}$

From $K_{3} PO_{4}$ : $0.4 m o l K_{3} PO_{4} \times \frac{1 m o l Mg _{3} ( PO _{4} ) _{2}}{2 m o l K _{3} PO _{4}} = 0.2 m o l Mg_{3} (PO_{4})_{2}$

Since both reactants yield $0.2$ moles of product, the maximum number of moles of magnesium phosphate is $0.2$ .

Chemistry - Mole Concept Question with Solution | TestHub

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