How many moles of [tex]Pb \left( NO_3 \right)_2[/tex] are required to generate 12 moles of [tex]Al \left( NO_3 \right)_3[/tex]?

[tex]\[
3 Pb \left( NO_3 \right)_2 + 2 AlCl_3 \rightarrow 3 PbCl_2 + 2 Al \left( NO_3 \right)_3
\][/tex]

[tex]\boxed{?}[/tex] mol [tex]Pb \left( NO_3 \right)_2[/tex]



Answer :

To find the number of moles of [tex]\( \text{Pb}( \text{NO}_3)_2 \)[/tex] required to produce 12 moles of [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex], let's follow these step-by-step instructions:

1. Identify the given data:
- The balanced chemical equation is:
[tex]\[ 3 \text{Pb}( \text{NO}_3)_2 + 2 \text{AlCl}_3 \rightarrow 3 \text{PbCl}_2 + 2 \text{Al}( \text{NO}_3)_3 \][/tex]
- We are given that 12 moles of [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex] are produced.

2. Determine the stoichiometric relationship:
- From the balanced equation, 3 moles of [tex]\( \text{Pb}( \text{NO}_3)_2 \)[/tex] produce 2 moles of [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex].

3. Set up a ratio to find the moles required:
- The molar ratio between [tex]\( \text{Pb}( \text{NO}_3)_2 \)[/tex] and [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex] is:
[tex]\[ \frac{3 \text{ moles } \text{ of } \text{Pb}( \text{NO}_3)_2}{2 \text{ moles } \text{ of } \text{Al}( \text{NO}_3)_3} \][/tex]
- This can be simplified as:
[tex]\[ \frac{3}{2} = 1.5 \][/tex]

4. Calculate the moles of [tex]\( \text{Pb}( \text{NO}_3)_2 \)[/tex] needed:
- To generate 12 moles of [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex]:
[tex]\[ \text{Moles of } \text{Pb}( \text{NO}_3)_2 = 12 \text{ moles } \text{ of } \text{Al}( \text{NO}_3)_3 \times \left( \frac{3 \text{ moles } \text{ of } \text{Pb}( \text{NO}_3)_2}{2 \text{ moles } \text{ of } \text{Al}( \text{NO}_3)_3} \right) \][/tex]
- Simplify the calculation:
[tex]\[ \text{Moles of } \text{Pb}( \text{NO}_3)_2 = 12 \times 1.5 = 18 \text{ moles} \][/tex]

Therefore, 18 moles of [tex]\( \text{Pb}( \text{NO}_3)_2 \)[/tex] are required to generate 12 moles of [tex]\( \text{Al}( \text{NO}_3)_3 \)[/tex].