Question

If 1.0 volumes of 1.0 M solutions of sodium hydroxide and lead (II) nitrate are mized how many moles of product are produced

Answer 1

Answer:

0.5 moles of Pb(OH)₂ are produced.

Explanation:    

The reaction is:  

$2NaOH + Pb(NO_{3})_{2} \rightarrow Pb(OH)_{2} + 2 NaNO_{3}$

If we have 1.0 liter of 1.0 M of sodium hydroxide and lead (II) nitrate, the number of moles are:  

$n_{NaOH} = C*V = 1 M*1 L = 1 mol$

$n_{Pb(NO_{3})_{2}} = C*V = 1 M*1 L = 1 mol$

Now, we need to find the limiting reactant knowing that 2 moles of sodium hydroxide react with 1 mol of lead (II) nitrate:

$n_{NaOH} = \frac{2 moles_{NaOH}}{1 mol Pb(NO_{3})_{2}}* 1 mol Pb(NO_{3})_{2} = 2 moles$

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:

$n_{Pb(OH)_{2}} = \frac{1 mol Pb(OH)_{2}}{2 mol NaOH}*1 mol NaOH = 0.5 mol$

Therefore, 0.5 moles of Pb(OH)₂ are produced.

I hope it helps you!