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Standard XII

Chemistry

Common Ion Effect

Which is the ...

Question

Which is the limiting reactant of the reaction that takes place in the solution?

P b (I O_{3})_{2}

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P b (N O_{3})_{2}

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K I O_{3}

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Both options B and C

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Solution

The correct option is C $P b (N O_{3})_{2}$
$2 K I O_{3} + P b (N O_{3})_{2} ⟶ P b (I O_{3})_{2} + 2 K N O_{3}$
$0.8 \times 15$ $35 \times 0.15$
$= 12$ $= 5.25$
$1$ mol of $P b (N O_{3})_{2}$ reacts with $2$ mol of $K I O_{3}$
$5.25$ mmol of $P b (N O_{3})_{2}$ reacts $= 2 \times 5.25 = 10.5$ mmol
The number of mmol of $K I O_{3}$ left $= 12 - 10.5 = 1.5$

Hence, $P b (N O_{3})_{2}$ is the limiting reactant.

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Similar questions

P b (I O_{3})_{2}

is a springly soluble salt

(K_{s p} = 2.6 \times 10^{- 13})

. To 35 mL of 0.15 M

P b (N O_{3})_{2}

solution, 15 mL of 0.8 M

K I O_{3}

solution is added, and a precipitate of

P b (I O_{3})_{2}

is formed.

What will be the molarity of

I O_{3}^{⊝}

ions in the solution after completion of the reaction?

Q. Consider the following balanced reaction:

2 C a (s) + O_{2} (g) \to 2 C a O

If 60 g of Ca is placed in a reaction chamber with 32 g of

O_{2},

which will be the limiting reactant and mass of the excess reactant left?

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2 A + 4 B \to 3 C + 4 D,$ when 5 moles of A react with 6 moles of B, then

(i) Which is the limiting reagent ?

(ii) Calculate the amount of C formed ?

C a {C l}_{2} + {B e}_{2} {C O}_{3} \to C a {C O}_{3} + B e {C l}_{3}

I

. Provided the reaction above, if

1.10 g

C a {C l}_{2}

and

0.85 g

{B e}_{2} {C O}_{3}

will be the limiting reactant.

I I

. The limiting reactant in a reaction will always be the reactant that has the lowest mass.

Q. Consider the following balanced reaction:

2 C a (s) + O_{2} (g) \to 2 C a O

If 60 g of Ca is placed in a reaction chamber with 32 g of

O_{2},

which will be the limiting reactant and mass of the excess reactant left?

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Which is the limiting reactant of the reaction that takes place in the solution?

The correct option is C Pb(NO3)22KIO3+Pb(NO3)2⟶Pb(IO3)2+2KNO30.8×15 35×0.15 =12 =5.251 mol of Pb(NO3)2 reacts with 2 mol of KIO35.25 mmol of Pb(NO3)2 reacts =2×5.25=10.5 mmolThe number of mmol of KIO3 left =12−10.5=1.5Hence, Pb(NO3)2 is the limiting reactant.