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------------------------------------------------ According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb(NO3)2 solution?

\[ 2 \text{KCl(aq)} + \text{Pb(NO}_3\text{)}_2 \text{(aq)} \rightarrow \text{PbCl}_2\text{(s)} + 2 \text{KNO}_3\text{(aq)} \]

A. 58.1 mL
B. 97.4 mL
C. 86.1 mL
D. 43.0 mL
E. 116 mL

Answer :

The volume of the 0.244 M KCl solution required to react exactly with 50.0 mL of 0.210 M Pb(NO₃)₂ solution is 86.1 mL.

The correct answer is C) 86.1 mL.

To determine the volume of 0.244 M KCl solution required to react with 50.0 mL of 0.210 M Pb(NO₃)₂ solution, we need to use the stoichiometry of the balanced equation.

The balanced equation is:

2 KCl(aq) +Pb(NO₃)₂ (aq) → PbCl₂(s) + 2 KNO₃(aq)

From the balanced equation, we can see that the molar ratio between KCl and Pb(NO₃)₂ is 2 ratio 1. This means that 2 moles of KCl react with 1 mole of Pb(NO₃)₂ .

First, we need to calculate the number of moles of Pb(NO₃)₂ in the given 50.0 mL of 0.210 M solution:

Moles of Pb(NO₃)₂ = Volume (L) × Concentration (M)

= 0.050 L × 0.210 M

= 0.0105 moles

Since the stoichiometric ratio between Pb(NO₃)₂ and KCl is 1 ratio 2, we can say that 0.0105 moles of Pb(NO₃)₂ will react with 2 × 0.0105 = 0.0210 moles of KCl.

Next, we can calculate the volume of 0.244 M KCl solution needed to provide 0.0210 moles:

Volume (L) = Moles / Concentration

= 0.0210 moles / 0.244 M

= 0.0861 L

Finally, we need to convert the volume to milliliters:

Volume (mL) = 0.0861 L × 1000 mL/L

= 86.1 mL

Therefore, the volume of the 0.244 M KCl solution required to react exactly with 50.0 mL of 0.210 M Pb(NO₃)₂ solution is 86.1 mL.

The correct answer is C) 86.1 mL.

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