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Limiting Reagent or Reactant

The addition ...

Question

The addition of 0.15 moles of $N a_{2} S O_{4}$ to an aqueous solution containing 0.05 moles of $B a C l_{2}$ results in the formation of an insoluble white precipitate of $B a S O_{4}$ . Find the limiting reagent and the amount of $B a S O_{4}$ that gets formed.

N a_{2} S O_{4}

, 11.65 g

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B a C l_{2}

, 34.95 g

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N a_{2} S O_{4}

, 34.95 g

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B a C l_{2}

, 11.65 g

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Solution

The correct option is D $B a C l_{2}$ , 11.65 g
$N a_{2} S O_{4} + B a C l_{2} \to 2 N a C l + B a S O_{4}$
Moles of $N a_{2} S O_{4}$ = 0.15 moles
Moles of $B a C l_{2}$ = 0.05 moles
1 mole of $N a_{2} S O_{4}$ reacts with 1 mole of $B a C l_{2}$
0.15 moles of $N a_{2} S O_{4}$ will react with 0.15 moles of $B a C l_{2}$
So, $B a C l_{2}$ is the limiting reagent.
As per the reaction, 1 mole of $B a C l_{2}$ produces 1 mole of $B a S O_{4}$
0.05 moles of $B a C l_{2}$ will produce 0.05 moles of $B a S O_{4}$
Amount of precipitate formed = moles of $B a S O_{4}$ formed $\times$ Molar mass of $B a S O_{4}$
Amount of precipitate formed = $0.05 \times 233 = 11.65$ g

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Limiting Reagent or Reactant

Standard XII Chemistry

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The addition of 0.15 moles of Na2SO4 to an aqueous solution containing 0.05 moles of BaCl2 results in the formation of an insoluble white precipitate of BaSO4. Find the limiting reagent and the amount ofBaSO4 that gets formed.

The addition of 0.15 moles of $N a_{2} S O_{4}$ to an aqueous solution containing 0.05 moles of $B a C l_{2}$ results in the formation of an insoluble white precipitate of $B a S O_{4}$ . Find the limiting reagent and the amount of $B a S O_{4}$ that gets formed.