Question

Question #a8585

Answer 1

Part One:
`2AlPO_4(aq)+3Mg(s)rarr2Al(s)+Mg_3(PO_4)_2(aq)`

Part Two:
Theoretical Yield of Aluminium = 1.430-g

Part Three:
Actual Yield of Aluminum = 1.6-g

Explanation:

Part One: Balance the Equation

_ `AlPO_4(aq)+` `Mg(s)rarr` `Al(s)+`_ `Mg_3(PO_4)_2(aq)`

The blank spaces in the reaction above represent the coefficients that will need to be added in order to balance the equation.
The goal is to add coefficients so that both sides of the equation will have equal numbers of elements.

Note that the product side of the equation has two `PO_4^(3-)` ions, but the reactant side only has one `PO_4^(3-)` ion. Assign a coefficient of 2 to `AlPO_4(aq)` so that both sides of the equation will have two `PO_4^(3-)` ions.

`2AlPO_4(aq)+` `Mg(s)rarr` `Al(s)+`_ `Mg_3(PO_4)_2(aq)`

Now balance `Al` by assigning ga coefficient of 2 to `Al(s)` since the reactant side has two `Al^(3+)`.

`2AlPO_4(aq)+` `Mg(s)rarr2Al(s)+` `Mg_3(PO_4)_2(aq)`

Finally, assign a coefficient of 3 to `Mg(s)` to match the three `Mg^(2+)` ions.

`2AlPO_4(aq)+3Mg(s)rarr2Al(s)+`_ `Mg_3(PO_4)_2(aq)`

Final Balanced Equation:
`2AlPO_4(aq)+3Mg(s)rarr2Al(s)+Mg_3(PO_4)_2(aq)`

NOTE: Both the reactant and product sides of the equation have 2 Al, 2 P, 8 O, and 3 Mg.

Part Two: Solve for Theoretical Yield of Aluminum

Given:
`m_(Mg(s))=1.932`-g

Solve for the theoretical yield of `Al(s)` in grams using mole ratios and molar masses.

`1.932_gMg*((1_(mol)Mg)/(24.305_gMg))*((2_(mol)Al)/(3_(mol)Mg))*((26.982_gAl)/(1_(mol)Al))=1.430_gAl`

Theoretical Yield = 1.430-g Al

Part Three: Solve for Actual Yield of Aluminum

Given:
`m_(Mg(s))=2.306`-g
`%Yield=0.96`

Begin by solving for the theoretical yield of `Al(s)` in grams using mole ratios and molar mass.

`2.306_gMg*((1_(mol)Mg)/(24.305_gMg))*((2_(mol)Al)/(3_(mol)Mg))*((26.982_gAl)/(1_(mol)Al))=1.707_gAl`

Now solve for the actual yield of `Al(s)`.

[Percent Yield (%)]`=`[Actual Yield (g)]`-:`[Theoretical Yield (g)]

[Actual Yield (g)]`=`[Theoretical Yield (g)]`*`[Percent Yield (%)]

`1.707_g*0.96=1.6_g`

1.6-g of aluminum metal should be obtained.