A mixture of MgBr2 and inert material is analyzed to determine the Mg content. First the...

A mixture of MgBr2 and inert material is analyzed to determine the Mg content. First the mixture is dissolved in water. Then all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. MgBr2(aq) + 2AgNO3(aq) 2AgBr(s) + Mg(NO3)2(aq) In one experiment, a 0.8107 g sample of the mixture resulted in 1.1739 g of AgBr. Determine the percent (by mass) of Mg in the mixture.

Expert Solution

moles of AgBr formed = mass / molar mass

= 1.1739 / 187.8

= 6.25 x 10^-3

MgBr2(aq) + 2AgNO3(aq)-----------------------> 2AgBr(s) + Mg(NO3)2(aq)

1 2

? 6.25 x 10^-3

MgBr2 moles = 6.25 x 10^-3 / 2 = 3.12 x 10^-3

moles of Mg = moles of MgBr2 = 3.12 x 10^-3

mass of Mg = moles x molar mass

= 3.12 x 10^-3 x 24

= 0.07501 g

mass percent of Mg = 0.07501 x 100 / 0.8107

= 9.25 %

queen_honey_blossom answered 2 years ago

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