Question

When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ × \[{10^{-5}}\] moles of lead sulphate precipitate out. (Round off to the nearest integer).

Answer 1

Hint: In a chemical reaction, the limiting reagent is the reactant that is present less in amount and initially gets consumed. This restricts the quantity of yield generated.

Formula Used:
Molarity = \[\dfrac{{{\rm{No}}{\rm{.of moles of solute}}}}{{{\rm{Volume of solution in litres}}}}\]

Complete Step by Step Solution:
Several times, reactions are accomplished with the quantities of reactants that are distinct from the quantities expected by a balanced chemical reaction.

In such conditions, one reactant is more surplus than the quantity expected by a balanced chemical reaction.

The reactant which is present in the limited quantity gets depleted after some time and after that further reaction does not take place whatever might be the quantity of the other reactant.

Hence, the reactant, which gets consumed at the initial stage, restricts the amount of product formed and is called the limiting reagent.

Lead nitrate solution reacts with chromic sulphate to form lead sulphate and chromic nitrate.

The balanced chemical equation is: \[{\rm{3Pb}}{\left({{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{+C}}{{\rm{r}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_4})_3}\to{\rm{3PbS}}{{\rm{O}}_{\rm{4}}}{\rm{+2Cr(N}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\]

Moles of lead nitrate
Molarity = 0.15 M = 0.15 moles per Litre
Volume = 35 mL
0.15 mol/L = no.of moles (\[\dfrac{{{\rm{1000}}}}{{{\rm{35L}}}}\])
\[ \Rightarrow \] No.of moles = (0.15) mol/L (\[\dfrac{{{\rm{35L}}}}{{{\rm{1000}}}}\])
\[ \Rightarrow \] No.of moles = 0.00525 = \[5.25 \times {10^{ - 3}}\]

Moles of lead nitrate
Molarity = 0.12 M = 0.12 moles per Litre
Volume = 20 mL
0.12 mol/L = no.of moles (\[\dfrac{{{\rm{1000}}}}{{{\rm{20L}}}}\])
\[ \Rightarrow \] No.of moles = (0.12) mol/L (\[\dfrac{{{\rm{20L}}}}{{{\rm{1000}}}}\])
 No.of moles = 0.0024 = .\[2.4 \times {10^{ - 3}}\].

If we compare the no.of moles of lead nitrate and chromic sulphate we can see that chromic sulphate is the limiting reagent, in this case.

So, lead sulphate would be formed only from the amount of available chromic sulphate i.e., 2.4×10-3 moles.
From the balanced chemical equation, we see that
1 mole of chromic sulphate gives 3 moles of lead sulphate.
\[2.4 \times {10^{ - 3}}\] moles of chromic sulphate will give 3(\[2.4 \times {10^{ - 3}}\]) moles of lead sulphate.

So,\[7.2 \times {10^{ - 3}}\]moles of lead sulphate will be formed.

Hence, \[0.07 \times {10^{ - 5}}\]moles of lead sulphate are formed.

Note: While attempting the question, limiting reagents must be kept in mind. From the limiting reagent, we can calculate the no.of moles of the product formed. Here the reactant which is present less in amount will act as the limiting reagent. Therefore, chromic sulphate is the limiting agent.