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Upon mixing 50.0 mL of 0.1 M lead nitrat...

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? Also, calculate the molar concentration of the species left behind in the final solution. Which is the limiting reagent?

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To solve the problem step by step, we will follow these steps: ### Step 1: Write the Balanced Chemical Equation The reaction between lead nitrate and chromic sulfate can be represented as: \[ 3 \text{Pb(NO}_3\text{)}_2 + \text{Cr}_2(\text{SO}_4\text{)}_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3\text{)}_3 \] ...
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