On to the Lab 5) This is one of my favorite demonstrations for stoichiometry. Al

Question

On to the Lab 5) This is one of my favorite demonstrations for stoichiometry. Aluminum metal reacts with Cuclh 2H 0(copperl) chloride dihydrate) to form copper metal, aluminum chloride and water. The Cuclh 2H20 must be dissolved in water first, as it is a solid. a Write the balanced chemical equation from the word equation given above, including all phase labels. Align your symbols and formulas with the pictures. H2O b) Given that 0.50 g Al and 5.00 g copper(ll) chloride dibvdrate are combined with 100.0 ml of water, determine which reactant is the limiting reactant and calculate the number of grams of copper metal that can form? c) Which reactant was oxidized? Explain how you could tell. Calculate the molarity of the aluminum chloride in the solution, assuming it all dissolved. Neglect the additional water added to the volume from the copper(l) chloride dibvdrate d)

Explanation / Answer

A)

balanced equation

Al(s) + CuCl2*2H2O(aq) = Cu(s) + AlCl3(aq) + H2O(l)

2Al(s) + 3CuCl2*2H2O(aq) = 3Cu(s) + 2AlCl3(aq) + 6H2O(l)

b)

mol of Al = mass/MW = 0.5/26.98 = 0.01853

mol of CuCl2*2H2O = mass/MW = (5)/(134.45 +18*2) = 0.02933

ratio is 2:3 so,

CuCl2 is in excess, Aluminium metal is limiting

0.01853 mol of Al + 3/2*0.01853 = 0.02779 mol of hydrate react

0.01853 mol of Al --> 3/2*0.01853 = 0.02779 mol of Copper (solid)

mass = molMW= 0.02779 *63.5 = 1.7646 g of Copper

c)

oxidation --> loss of e-, this must be aluminium

Al goes from 0 in Al(s) to +3 in AlCl3(aq)

d)

mol of AlCl3 produced = 0.01853

[AlCl3] = mol/V = (0.01853)/(0.1) = 0.1853 M

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.