Question

1. beaker x and beaker y each contain 1.0 l of solution, as shown above. a student combines the solutions by pouring them into a larger, previously empty beaker z and observes the formation of a white precipitate. assuming that volumes are additive, which of the following sets of solutions could be represented by the diagram above? (a) beaker x: 2.0 m agno₃, beaker y: 2.0 m mgcl₂, beaker z: 4.0 m mg(no₃)₂ and agcl(s) (b) beaker x: 2.0 m agno₃, beaker y: 2.0 m mgcl₂, beaker z: 2.0 m mg(no₃)₂ and agcl(s) (c) beaker x: 2.0 m agno₃, beaker y: 1.0 m mgcl₂, beaker z: 1.0 m mg(no₃)₂ and agcl(s) (d) beaker x: 2.0 m agno₃, beaker y: 1.0 m mgcl₂, beaker z: 0.50 m mg(no₃)₂ and agcl(s)

Explanation:

Step1: Write the chemical reaction

$$2AgNO_{3}(aq)+MgCl_{2}(aq)=2AgCl(s)+Mg(NO_{3})_{2}(aq)$$

Step2: Calculate the moles of reactants

For \(AgNO_{3}\) in beaker \(X\): \(n(AgNO_{3}) = M\times V=2.0\ M\times1.0\ L = 2.0\ mol\)
For \(MgCl_{2}\) in beaker \(Y\) (assuming option \(D\) where \(M = 1.0\ M\) and \(V = 1.0\ L\)): \(n(MgCl_{2})=1.0\ M\times1.0\ L = 1.0\ mol\)

Step3: Determine the limiting reactant

From the stoichiometry (\(2\ mol\ AgNO_{3}\) reacts with \(1\ mol\ MgCl_{2}\)), they react completely.

Step4: Calculate the concentration of \(Mg(NO_{3})_{2}\) in beaker \(Z\)

Total volume \(V_{total}=1.0\ L + 1.0\ L=2.0\ L\)
Moles of \(Mg(NO_{3})_{2}\) produced: \(n(Mg(NO_{3})_{2}) = 1.0\ mol\) (from \(MgCl_{2}\) moles, since \(1\ mol\ MgCl_{2}\) gives \(1\ mol\ Mg(NO_{3})_{2}\) )
Concentration \(M=\frac{n}{V}=\frac{1.0\ mol}{2.0\ L}=0.50\ M\)

Answer:

D. \(2.0\ M\ AgNO_{3}\), \(1.0\ M\ MgCl_{2}\), \(0.50\ M\ Mg(NO_{3})_{2}\) and \(AgCl(s)\)