Question & Answer: Gerardo dissolves 12.65 g of solid Al(NO_3)_3 in enough water to make 200.0 mL…..

2) Gerardo dissolves 12.65 g of solid Al(NO3)3 in enough water to make 200.0 mL of solution. Marciela then adds enough solid Mg(NO3)2 to increase the concentration of nitrate ions to 1.000 M. Assuming that the solution volume does not change significantly, what mass of Mg(NO3)2 did Marciela add?

Gerardo dissolves 12.65 g of solid Al(NO_3)_3 in enough water to make 200.0 mL of solution. Marciela then adds enough solid Mg(NO_3)_2 to increase the concentration of nitrate ions to 1.000 M. Assuming that the solution volume does not change significantly, what mass of Mg(NO_3)_2 did Marciela add?

Expert Answer

Answer

Moles of Al(NO3)3 = 12.65/213 = 0.0594 moles

Molarity of Al(NO3)3 = 0.0594/0.200 = 0.3 M

The total moles of NO3- required = 0.200 L x 1.0 mol/L

= 0.200 mol

The moles of NO3- provided by the Al(NO3)3
= 3 mol NO3-/1mol Al(NO3)3 x 0.200 L x 0.3 M Al(NO3)3

= 0.180 mol NO3-

Moles of NO3- to be provided by the Mg(NO3)2 would be = Total moles NO3- – moles NO3- from Al(NO3)3

= 0.200 – 0.180

= 0.02 mol NO3-

Since 1 mol Mg(NO3)2 provides 2 mol of NO3-, the moles of Mg(NO3)2 required would be 1/2 the moles of NO3-, so

mol Mg(NO3)2 = 1 mol Mg(NO3)2/ 2 mol NO3- x 0.02 mol NO3-= 0.01 mol Mg(NO3)2

Convert moles of Mg(NO3)2 to grams using the mole mass

Mass of Mg(NO3)2 = 0.01 * 148.3 = 1.48 g

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