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, show your work so i understand how to do this. use whole numbers for the molar masses and give answers to 3 decimal places.

I got a different answer.

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

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