Upon mixing `50.0 mL` of `0.1 M` lead nitrate solution with `50.0 mL` of `0.05 M` chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed and what is the molar concertration of chromic suplhate left in the solution?

Options

A. 0.0005,0.0084
B. 0.0084,0.005
C. 0.005,0.00084
D. 0.05,0.00084

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced chemical equation
The reaction between lead nitrate and chromic sulfate can be represented as follows:

\[ 3 \text{Pb(NO}_3\text{)}_2 + \text{Cr}_2(\text{SO}_4\text{)}_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3\text{)}_3 \]

### Step 2: Calculate the moles of lead nitrate and chromic sulfate
- For lead nitrate (\(\text{Pb(NO}_3\text{)}_2\)):
\[ \text{Moles of Pb(NO}_3\text{)}_2 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} = 5 \, \text{mmoles} \]

- For chromic sulfate (\(\text{Cr}_2(\text{SO}_4\text{)}_3\)):
\[ \text{Moles of Cr}_2(\text{SO}_4\text{)}_3 = 0.05 \, \text{M} \times 0.050 \, \text{L} = 0.0025 \, \text{moles} = 2.5 \, \text{mmoles} \]

### Step 3: Determine the limiting reagent
From the balanced equation, we see that:
- 3 mmoles of \(\text{Pb(NO}_3\text{)}_2\) react with 1 mmole of \(\text{Cr}_2(\text{SO}_4\text{)}_3\).

To find out how much \(\text{Cr}_2(\text{SO}_4\text{)}_3\) is required for 5 mmoles of \(\text{Pb(NO}_3\text{)}_2\):
\[ \text{Required moles of Cr}_2(\text{SO}_4\text{)}_3 = \frac{1}{3} \times 5 = 1.67 \, \text{mmoles} \]

Since we have 2.5 mmoles of \(\text{Cr}_2(\text{SO}_4\text{)}_3\) available, \(\text{Pb(NO}_3\text{)}_2\) is the limiting reagent.

### Step 4: Calculate the moles of lead sulfate formed
According to the stoichiometry of the reaction:
- 3 mmoles of \(\text{Pb(NO}_3\text{)}_2\) produce 3 mmoles of \(\text{PbSO}_4\).

Thus, 5 mmoles of \(\text{Pb(NO}_3\text{)}_2\) will produce:
\[ \text{Moles of PbSO}_4 = 5 \, \text{mmoles} = 0.005 \, \text{moles} \]

### Step 5: Calculate the remaining moles of chromic sulfate
Initial moles of \(\text{Cr}_2(\text{SO}_4\text{)}_3\) = 2.5 mmoles
Moles of \(\text{Cr}_2(\text{SO}_4\text{)}_3\) consumed:
\[ \text{Consumed moles} = \frac{1}{3} \times 5 = 1.67 \, \text{mmoles} \]
Remaining moles:
\[ \text{Remaining moles} = 2.5 - 1.67 = 0.83 \, \text{mmoles} \]

### Step 6: Calculate the molar concentration of chromic sulfate left in the solution
The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

The concentration of remaining \(\text{Cr}_2(\text{SO}_4\text{)}_3\) is:
\[ \text{Concentration} = \frac{\text{Remaining moles}}{\text{Total volume}} = \frac{0.00083 \, \text{moles}}{0.1 \, \text{L}} = 0.0083 \, \text{M} \]

### Final Answers
- Moles of lead sulfate formed: **0.005 moles**
- Molar concentration of chromic sulfate left in the solution: **0.0083 M**