Answer to Question #202386 in General Chemistry for jordan

Question

4. If you had 36.2 grams of aluminum metal reaction with 45.0 grams of copper (II) hydroxide,

which is the limiting and which is the excess reactant?

5. How many grams of copper metal would be produced?

6.How much of the excess reactant is left when the reaction stops?

Answer 1

3Cu(OH)₂ + 2Al → 3Cu + 2Al(OH)₃

$Moles (Al)$ $= \frac{36.2}{26.98}=1.34moles$

$Moles[Cu(OH)_2]=\frac{45.0}{97.561}=0.46moles$

Aluminum is the limiting and copper (ii) hydroxide is the excess reactant.

Moles of copper metal = 0.46 moles

Mass of copper metal= 0.46×3=1.38