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Chapter 16, Problem 106P

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QUESTION:

Problem 106P

Determine whether HCl can dissolve each metal sample. If it can, write a balanced chemical reaction showing how the metal dissolves in HCl and determine the minimum amount of 6.0 M HCl required to completely dissolve the sample.

(a) 5.90 g Ag

(b) 2.55 g Pb

(c) 4.83 g Sn

(d) 1.25 g Mg

Questions & Answers

QUESTION:

Problem 106P

Determine whether HCl can dissolve each metal sample. If it can, write a balanced chemical reaction showing how the metal dissolves in HCl and determine the minimum amount of 6.0 M HCl required to completely dissolve the sample.

(a) 5.90 g Ag

(b) 2.55 g Pb

(c) 4.83 g Sn

(d) 1.25 g Mg

ANSWER:

Solution 106P:

Here, we are going to determine whether HCl can dissolve the given metal samples and write a balanced chemical equation for the ones which can be dissolved.

Step1:

To determine whether HCl can dissolve the given metals, we have to check the position of the metals in the activity series. The activity series is a list of metals arranged in the order of their decreasing activities.

                

Step2:

As we have seen. Out of all the metals given in the problem, only silver(Ag) is placed below hydrogen in the activity series. Thus, only Ag will not be dissolved in HCl. Lead(Pb), Tin(Sn) and Magnesium(Mg) will be dissolved in HCl.  The balanced equations involved in the process are:

                        Pb(s) + 2HCl(aq) → PbCl2(aq) + H2(g)

Sn(s) + 2HCl(aq) → SnCl2(aq) + H2(g)

                        Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step3:

Determination of amount of HCl required to dissolve 2.55 g Pb:

Given,

Mass of Pb used = 2.55 g

Molar mass of Pb = 207.2 g/mol

Therefore, number of moles of Pb = given mass / molar mass

                                       = 2.55 g / (207.2 g/mol)

                                       = 0.0123 mol

Step4:

From the above balanced equation, it is clear that,

To dissolve 1 mol of Pb, number of moles of HCl required = 2 mol

Therefore, to dissolve 0.0123 mol of Pb, number of moles of HCl required = 2 x 0.0123

                                                                                       = 0.0246 mol

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