Answer:
0.5 moles of Pb(OH)₂ are produced.
Explanation:
The reaction is:
[tex]2NaOH + Pb(NO_{3})_{2} \rightarrow Pb(OH)_{2} + 2 NaNO_{3}[/tex]
If we have 1.0 liter of 1.0 M of sodium hydroxide and lead (II) nitrate, the number of moles are:
[tex] n_{NaOH} = C*V = 1 M*1 L = 1 mol [/tex]
[tex] n_{Pb(NO_{3})_{2}} = C*V = 1 M*1 L = 1 mol [/tex]
Now, we need to find the limiting reactant knowing that 2 moles of sodium hydroxide react with 1 mol of lead (II) nitrate:
[tex] n_{NaOH} = \frac{2 moles_{NaOH}}{1 mol Pb(NO_{3})_{2}}* 1 mol Pb(NO_{3})_{2} = 2 moles [/tex]
Since we have 1 mol of sodium hydroxide and we need 2 moles to react with lead (II) nitrate, then the limiting reactant is sodium hydroxide.
We can find the number of moles of lead (II) hydroxide produced as follows:
[tex] n_{Pb(OH)_{2}} = \frac{1 mol Pb(OH)_{2}}{2 mol NaOH}*1 mol NaOH = 0.5 mol [/tex]
Therefore, 0.5 moles of Pb(OH)₂ are produced.
I hope it helps you!
