Question

Al(s)+ CuSO₄ --> Al(SO₄)₃ + Cu


If 10 grams of Aluminum are used with 75 mL of a 2.0M copper (II) sulfate solution, how much copper will be theoretically produced?

1. Convert 10 grams of aluminum to moles. ___________

2. Convert 75 mL of 2.0M copper (II) sulfate to moles. __________

3. Convert moles of aluminum used to moles of copper produced using the Balanced Equation Ratios: _______ moles Cu

4. Convert moles of copper to grams of copper: _______ grams Cu


(I got 0.371 , 0.150, 3, and 127.09) Are these correct? If not, please show your work so I understand how to do this. Use whole numbers for the molar masses and give answers to 3 decimal places.

Answer 1

I got a different answer.

Answer 2

One

The molar mass of aluminum from the periodic table is 27

moles = given mass / molar mass

moles = 10 / 27

moles = 0.3704

Two

75 mL * [1 L / 1000 mL] = 0.075 mL

Concentration = 2M = 2 moles / L

Concentration = moles / Volume

2 = moles / 0.075 L        Multiply both sides by 0.075

2*0.075 = moles

0.150 moles

Three

Balance Equation: 2Al(s)+ 3CuSO₄ --> Al_2(SO₄)₃ + 3Cu   Note the formula for Al_2 (SO_4)_3

2 Al / 0.3704 moles = 3Cu / x     Substitute

2 / 0.3704 = 3/x                            Cross Multiply

2x = 0.3704 * 3

2x = 1.111                                         Divide by 2

x = 1.111/2

x = 0.555555 moles. of Copper

Four

moles = given mass / molar mass

molar mass Cu = 63.5

moles = 0.555

given mass = moles * molar mass

given mass = 35.28