Question

how many moles of lead (ii) hydroxide (solid) can be formed when 0.0225L of 0.135 M Pb(NO3)2 solution reacts with excess sodium hydroxide

Answer 1

0.0225L *0.135 M Pb(NO3)2=0.00304 mol Pb(NO3)2 given

Pb(NO3)2       +        2NaOH --->   Pb(OH)2+2NaNO3
0.00304 mol                                  0.00304mol
Answer 0.00304 mol Pb(OH)2

Answer 2

Answer : The number of moles of lead(II)hydroxide are, 0.00304 mole

Explanation : Given,

Molarity of $Pb(NO_3)_2$ = 0.135 M = 0.135 mole/L

Volume of solution = 0.0225 L

First we have to calculate the moles of $Pb(NO_3)_2$

$Molarity=\frac{\text{Moles of }Pb(NO_3)_2}{\text{volume of solution in liters}}$

Now put all the given values in this formula, we get the moles of $Pb(NO_3)_2$

$0.135mole/L=\frac{\text{Moles of }Pb(NO_3)_2}{0.0225L}$

$\text{Moles of }Pb(NO_3)_2}=0.00304mole$

Now we have to calculate the moles of lead(II)hydroxide.

The balanced chemical reaction will be,

$Pb(NO_3)_2+2NaOH\rightarrow Pb(OH)_2+2NaNO_3$

From the balanced chemical reaction, we conclude that

As, 1 mole of $Pb(NO_3)_2$ react to give 1 mole of $Pb(OH)_2$

So, 0.00304 mole of $Pb(NO_3)_2$ react to give 0.00304 mole of $Pb(OH)_2$

Therefore, the number of moles of lead(II)hydroxide are, 0.00304 mole