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Chapter 3: Q3.110P (page 89)

A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution:

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)

After 1.32 g of the impure metal was treated with 0.100 L of 0.750 M HCl, 0.0125 mol of HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the sample?

Short Answer

Expert verified

The mass percentage of Mg in the sample is 57.6 % Mg.

Step by step solution

01

Finding the moles of HCl used

Calculate the number of moles of HCl:

 Moles of HCl=0.1 L soln×0.75 mol HClL soln                                          =0.075 mol.Moles of HCl  used=0.0750 mol-0.0125 mol                                          =0.0625 mol.

02

Finding the mass percentage of Mg

On multiplying the mole number of HCl used by molar ratio, we get:

Moles of Mg=0.0625 mol HCl×1 mol Mg2 mol HCl                           =0.03125 mol  Mg.\hfillMass of Mg=0.03125 mol Mg×24.31 g Mg1 mol Mg                            =076 g.

Divide the mass of Ag to the mass of the impure metal, then multiply by 100:

Mass % of Mg=0.76 g Mg1.32 g impure metal×100                                =57.6%.

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