Question

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.

Answer 1

Answer:

9.24% the percent of Mg in the mixture.

Explanation:

Mass of silver bromide = 1.1739 g

Moles of silver bromide = $\frac{1.1739 g}{188 g/mol}=0.006244 mol$

1 mole of AgBr has 1 mole of bromine atom.Then 0.006244 moles AgBr has 0.006244 mole bromine atom.

Moles of bromine atoms = 0.006244 moles

Magnesium bromide has 2 moles of bromine atom.Then 0.006244 moles of bromine atoms will be in :

$\frac{0.006244 mol}{2}=0.003122 mol$

Magnesium bromide has 1 mole magnesium atom, then 0.003122 moles of magnesium bromide will have :

$1\times 0.003122 mol=0.003122 mol$ magnesium atom

Mass of 0.003122 moles of magnesium :

0.003122 mol × 24 g/mol = 0.07493 g

Mass of the sample = 0.8107 g

Percentage of magnesium in the sample ;

$=\frac{0.07493 g}{0.8107 g}\times 100=9.24\%$

9.24% the percent of Mg in the mixture.