Question

Consider the following chemical reaction $Pb(NO_3)_2 + Na_2SO_4 \rightarrow PbSO_4 + 2NaNO_3$ $\Rightarrow 85$ If a series of experiments run, maintaining sum of the weights of two reactant constant but varying the weights of reactants, which of the following statements is (are) not true ? $[M_{Pb(NO_3)_2} = 394, M_{Na_2SO_4} = 142]$

• A. Maximum weight of the ppt ($PbSO_4$) will be formed if equal weights of reactants are taken. X
• B. Maximum weight of the ppt ($PbSO_4$) will be formed if equal moles of reactants are taken X
• C. In the experiment as the weight of $Na_2SO_4$ increases, weight of ppt ($PbSO_4$) always increases X
• D. In the experiment as the weight of $Pb(NO_3)_2$ increases, weight of ppt ($PbSO_4$) always increases

Answer 1

A, C, D

Answer 2

The given chemical reaction is:

$Pb(NO_3)_2 + Na_2SO_4 \rightarrow PbSO_4 + 2NaNO_3$

The molar masses are:

$M_{Pb(NO_3)_2} = 394 \text{ g/mol}$ $M_{Na_2SO_4} = 142 \text{ g/mol}$

From the balanced chemical equation, 1 mole of $Pb(NO_3)_2$ reacts with 1 mole of $Na_2SO_4$ to produce 1 mole of $PbSO_4$.

Let $W_1$ be the weight of $Pb(NO_3)_2$ and $W_2$ be the weight of $Na_2SO_4$. The problem states that the sum of the weights of the two reactants is constant, let this be $W_{total}$. So, $W_1 + W_2 = W_{total}$ (constant).

To maximize the weight of $PbSO_4$ (the precipitate), both reactants should be consumed completely, which means they must be present in their stoichiometric molar ratio.

Stoichiometric molar ratio: $n_{Pb(NO_3)_2} : n_{Na_2SO_4} = 1:1$. This implies $n_1 = n_2$. In terms of mass, this means:

$\frac{W_1}{M_{Pb(NO_3)_2}} = \frac{W_2}{M_{Na_2SO_4}}$

$\frac{W_1}{394} = \frac{W_2}{142}$

So, the stoichiometric mass ratio is $W_1 : W_2 = 394 : 142$. This means for maximum $PbSO_4$ formation, $W_1$ should be approximately $2.77$ times $W_2$.

Let's find the specific weights for maximum product:

$W_1 = \frac{394}{142} W_2$

Substitute this into $W_1 + W_2 = W_{total}$:

$\frac{394}{142} W_2 + W_2 = W_{total}$

$W_2 \left(\frac{394 + 142}{142}\right) = W_{total}$

$W_2 \left(\frac{536}{142}\right) = W_{total}$

$W_{2,opt} = \frac{142}{536} W_{total} \approx 0.265 W_{total}$

Then, $W_{1,opt} = W_{total} - W_{2,opt} = W_{total} - \frac{142}{536} W_{total} = \frac{394}{536} W_{total} \approx 0.735 W_{total}$. These are the optimal weights for maximum $PbSO_4$ formation.

Now let's evaluate each statement:

(A) Maximum weight of the ppt ($PbSO_4$) will be formed if equal weights of reactants are taken.

If equal weights are taken, $W_1 = W_2 = W_{total}/2$. Comparing this with the optimal weights:

$W_{1,opt} \approx 0.735 W_{total}$ and $W_{2,opt} \approx 0.265 W_{total}$.

Since $W_{total}/2 \neq 0.735 W_{total}$ and $W_{total}/2 \neq 0.265 W_{total}$, equal weights do not correspond to the stoichiometric ratio by mass. Therefore, statement (A) is not true.

(B) Maximum weight of the ppt ($PbSO_4$) will be formed if equal moles of reactants are taken.

As established, equal moles ($n_1 = n_2$) is the condition for reactants to be in their stoichiometric ratio, leading to complete consumption of both reactants (if no other factors limit the reaction) and thus maximum product formation for a given total amount of reactants. Therefore, statement (B) is true.

(C) In the experiment as the weight of $Na_2SO_4$ increases, weight of ppt ($PbSO_4$) always increases.

Let $W_2$ be the weight of $Na_2SO_4$. Then $W_1 = W_{total} - W_2$. The amount of $PbSO_4$ formed depends on the limiting reactant.

If $W_2 < W_{2,opt}$ (i.e., $Na_2SO_4$ is limiting), the moles of $PbSO_4$ formed are $W_2/142$. In this region, as $W_2$ increases, the weight of $PbSO_4$ increases.

If $W_2 > W_{2,opt}$ (i.e., $Pb(NO_3)_2$ is limiting), the moles of $PbSO_4$ formed are $W_1/394 = (W_{total} - W_2)/394$. In this region, as $W_2$ increases, $(W_{total} - W_2)$ decreases, so the weight of $PbSO_4$ decreases.

Since the weight of $PbSO_4$ first increases and then decreases as $W_2$ increases, it does not always increase. Therefore, statement (C) is not true.

(D) In the experiment as the weight of $Pb(NO_3)_2$ increases, weight of ppt ($PbSO_4$) always increases.

Let $W_1$ be the weight of $Pb(NO_3)_2$. Then $W_2 = W_{total} - W_1$. The amount of $PbSO_4$ formed depends on the limiting reactant.

If $W_1 < W_{1,opt}$ (i.e., $Pb(NO_3)_2$ is limiting), the moles of $PbSO_4$ formed are $W_1/394$. In this region, as $W_1$ increases, the weight of $PbSO_4$ increases.

If $W_1 > W_{1,opt}$ (i.e., $Na_2SO_4$ is limiting), the moles of $PbSO_4$ formed are $W_2/142 = (W_{total} - W_1)/142$. In this region, as $W_1$ increases, $(W_{total} - W_1)$ decreases, so the weight of $PbSO_4$ decreases.

Since the weight of $PbSO_4$ first increases and then decreases as $W_1$ increases, it does not always increase. Therefore, statement (D) is not true.

The question asks for statements that are not true. Based on the analysis, statements (A), (C), and (D) are not true.

The final answer is $\boxed{A,C,D}$

Explanation of the solution:

The reaction requires $Pb(NO_3)_2$ and $Na_2SO_4$ in a 1:1 molar ratio. For maximum product ($PbSO_4$) when the total mass of reactants is constant, the reactants must be in this stoichiometric molar ratio. This translates to a mass ratio of $394:142$ ($M_{Pb(NO_3)_2}:M_{Na_2SO_4}$).

(A) Equal weights ($W_1=W_2$) do not match the required $394:142$ mass ratio, so this will not yield maximum product. Thus, (A) is false.

(B) Equal moles directly corresponds to the stoichiometric ratio, which ensures maximum product. Thus, (B) is true.

(C) and (D) describe varying reactant weights while keeping the total constant. The amount of product formed will increase as the limiting reactant's amount increases until the stoichiometric point is reached, after which the other reactant becomes limiting and further increase in the initial reactant's amount will lead to a decrease in product. Therefore, the product weight does not always increase with the increase of either reactant's weight. Thus, (C) and (D) are false.

Answer: A, C, D