Question

How many moles of precipitate will be formed when 100.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?

2 NaBr (aq) + Pb(NO₃)₂ (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)

Answer 1

Answer : The number of moles of precipitate, $PbBr_2$ formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

$\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}$

$\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol$

Now we have to calculate the moles of precipitate, $PbBr_2$ formed.

The balanced chemical reaction is:

$2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of $PbBr_2$

So, 0.02 moles of NaBr react to give $\frac{0.02}{2}=0.01$ mole of $PbBr_2$

Therefore, the number of moles of precipitate, $PbBr_2$ formed will be 0.01 moles.

Answer 2

The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole

• We'll begin by calculating the number of mole of NaBr in 100 mL of 0.2 M NaBr solution. This can be obtained as follow:

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaBr = 0.2 M

Mole of NaBr =?

Mole = Molarity x Volume

Mole of NaBr = 0.2 × 0.1

Mole of NaBr = 0.02 mole

• Finally, we shall determine the number of mole of the precipitate (i.e PbBr₂) produced from the reaction. This can be obtained as follow:

2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)

From the balanced equation above,

2 moles of NaBr reacted to produce 1 mole of PbBr₂.

Therefore,

0.02 mole of NaBr will react to produce = $\frac{0.02}{2}$ = 0.01 mole of PbBr₂.

Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole

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