Question

Stoichiometry

Stoichiometry (Mass to Mass) 2 Al (s)3 FeCl (aq)2 AIC, (ag)+3 Fe (s) 4. a. How many moles of iron are produced when 4.788 grams of aluminum completely reacts with iron (1) chloride? (0.2662 mol Fe) b. How many grams of aluminum chloride are formed when 1.50 moles of iron (1) chloride is consumed completely? (133 g AlCh c. How many grams of iron () chloride reacted to form 15.786 g of aluminum chloride? (22.509 g Feco,) Solutions and Solution Stoichiometry Naco, (s) → 2 Na. (aq) + CO-(aq) 5. What is the molarity of Na' ions in a 0.20 M Na CO, solution? (0.40 M) 84

Answer 1

a. The mass of aluminium= 4.788 g

The molar mass of aluminium = 27 g mol^-1

The no. of moles of aluminium = 4.788/27 = 0.177 mol

The no. of moles of iron produced per 2 moles of aluminium = 3 mol

Therefore, the no. of moles iron produced per 0.177 moles of aluminium = (0.177/2)*3 = 0.266 mol

b. The no. of moles AlCl₃ formed per 3 moles of FeCl₂ = 2 mol

Therefore, the no. of moles of AlCl₃ formed per 1.5 moles of FeCl₂ = (1.5/3)*2 = 1 mol

The molar mass of AlCl₃ = 133.5 g mol^-1.

Hence, the mass of AlCl₃ formed per 1.5 moles of FeCl₂ = 1 mol *133.5 g/mol = 133.5 g

c. The mass of AlCl₃ formed = 15.786 g

Therefore, from part b, the no. of moles of FeCl₂ per 15.786 g of AlCl₃ = (15.786/133.5)*1.5 = 0.177 mol

The molar mass of FeCl₂ = 126.75 g/mol

Hence, the no. of moles of FeCl₂ = 0.177 mol * 126.75 g/mol = 22.5 g