Question

Given the equation:
\[Pb{(S{O_4})_2} + 4LiN{O_3} \to Pb{(N{O_3})_4} + 2L{i_2}S{O_4}\]
How many grams of lithium nitrate will be needed to make 250 grams of lithium sulphate, assuming that you have an adequate amount of lead (IV) to do the reaction?

Answer 1

Hint:In order to find how many grams of lithium nitrate is needed to make 250 grams of lithium sulphate, we must first have an idea about what kind of reaction is taking place. The given reaction is a double displacement reaction. The double displacement is a reaction in which one component of each compound will be exchanged between one another.

Complete answer:
Let us first write the given balanced equation.
\[Pb{(S{O_4})_2} + 4LiN{O_3} \to Pb{(N{O_3})_4} + 2L{i_2}S{O_4}\]
The above given reaction is a double displacement reaction. The double displacement is a reaction in which one component of each compound will be exchanged between one another.
Double displacement cannot occur unless one of the products is an insoluble gas, solid precipitate or water. In the above equation, the product is none of these. All we have is a mixture of \[P{b^{4 + }}\], \[SO_4^{2 - }\], \[L{i^ + }\]and \[NO_3^ - \]ions.
Step 1: We might be requiring the molar mass of \[LiN{O_3}\]and \[L{i_2}S{O_4}\].
Molar mass of Lithium nitrate = \[68.9459gmo{l^{ - 1}}\]
Molar mass of lithium sulphate = \[109.9446gmo{l^{ - 1}}\]
This reaction will follow the following pattern which is given below
\[{\text{mass of product}} \to {\text{moles of product}} \to {\text{moles of reactant}} \to {\text{mass of reactant }}\]
Step 2: we have to convert the mass of \[L{i_2}S{O_4}\]to moles of \[L{i_2}S{O_4}\]
We have to divide the given mass of lithium sulphate with its molar mass.
\[250gL{i_2}S{O_4} \times \dfrac{{1molL{i_2}S{O_4}}}{{109.9446gL{i_2}S{O_4}}} = 2.27mol{\text{ }}L{i_2}S{O_4}\]
Step 3: now we have to convert moles of \[L{i_2}S{O_4}\]to moles of \[LiN{O_3}\]
In order to obtain the moles of \[LiN{O_3}\], we have to multiply the moles of \[L{i_2}S{O_4}\]with the mole dfraction between lithium nitrate and lithium sulphate from the equation
\[2.27mol{\text{ }}L{i_2}S{O_4} \times \dfrac{{2mol{\text{ }}LiN{O_3}}}{{4mol{\text{ }}L{i_2}S{O_4}}} = 4.548mol{\text{ }}LiN{O_3}\]
Step 4: Finally convert the moles of \[LiN{O_3}\]to mass of \[LiN{O_3}\]
So, we have to multiply the moles of \[LiN{O_3}\]with its molar mass.
\[4.548mol{\text{ }}LiN{O_3} \times \dfrac{{68.9459g{\text{ }}LiN{O_3}}}{{1mol{\text{ }}LiN{O_3}}} = 310g{\text{ }}LiN{O_3}\]
Therefore, 310g of lithium nitrate is needed to make 250 grams of lithium sulphate.

Note:
We have to note a certain point that the double displacement reaction is not a redox reaction but the displacement reaction is a redox reaction. Double displacement reaction is not redox because there is no change in the oxidation state of the compounds.