Aluminum and Silver Sulfate

By: Mason Prough

Proof of the reaction

Molar Mass for each reactant and product

Aluminium: Since there is only one Aluminium molecule on the left, you just use the atomic mass, which is 26.98 g/mol.

Silver: Silver has a subscript of 2 thanks to the sulfate. Now, you multiply the atomic mass of silver, 107.9 g/mol, by 2 making the result 215.8 g/mol.

Sulfur: There is only one sulfur molecule, so the atomic mass is the molar mass. 32.06 g/mol.

Oxygen: There are 4 oxygen molecules so, you multiply the atomic mass by 4 making 63.99 g/mol.

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Aluminum: This time, Aluminum has a subscript of 2. You just multiply the atomic mass by 2 making it 53.97 g/mol.

Sulfur: This part can be tricky, but it easy. The entire compound of Sulfate now has a subscript of 3 since Aluminum has a superscript of 3. This now makes 3 times the atomic mass which turns into 96.18 g/mol.

Oxygen: Like stated before, sulfate has a subscript of 3, so you multiply oxygen's atomic mass by 3, making it 191.9 g/mol.

Silver: Silver is all alone when replaced with Aluminum, so the molar mass is the atmoic mass of 107. 8 g/mol.

Mole to mole conversions

11.30 mol Al x 1 mol / 2 mol Al= 5.65 g/mol Al2(SO4)3


12.1 mol Ag x 3 Ag2(SO4) / 6 mol Ag = 6.05 mol Ag2(SO4)

Mass to Mass Conversion

3.29 g Al x 342.15 g Al2(SO4) / 26.26.9 g Al x 2 mol Al = 20.9 g Al2(SO4)

Limiting and Excess reactans

5.25 g Al x 6 mol Ag x 107.86 g Ag / 26.85 g Al x 2 mol Al = 62.9 g Ag

[ER]

12.18 g Ag2(SO)4 x 6 mol Ag x 107.87 g Ag / 311.82 g Ag2(SO4) x 3 mol Ag = 8.427 g Ag

[LR]

Usually the smaller answer is the Limiting Reactant

Theoretical Yield

This yield is usually the limiting reactant so in this case, it would be 8.427 g of Ag.

Percent yield

To calculate the Percent Yield, you have to use the equation, actual yield / theoretical yield x 100. For Ag or, Silver, you divide 8.400 (actual yield) by 8.427 (theoretical yield). Then when x by 100, you get 99.68%