1. Aluminum chloride (AlCl3), and sodium hydroxide (NaOH) can react to form aluminum hydroxide (Al (OH) 3) and sodium chloride (NaCl). You have 13.4 g of aluminum chloride and 10.0 g of sodium hydroxide.

Answer the following questions:

•What is the balanced equation for this reaction?

•If you use all 13.4 g of aluminum chloride, how many grams of aluminum hydroxide can be formed?

Work must be shown to earn credit

•If you use all 10.0 g of sodium hydroxide, how many grams of aluminum hydroxide can be formed? Work must be shown to earn credit

•How many grams of aluminum hydroxide will actually be made? Which reagent is limiting? Explain your answer.

1)

a) The balanced reaction between AlCl3 and NaOH is:

AlCl3 + 3NaOH → Al (OH) 3 + 3NaCl

b) Given mass of AlCl3 = 13.4 g

Molar mass of AlCl3 = 133.34 g/mol

# moles of AlCl3 = 13.4/133.34 = 0.1005 moles

Based on the reaction stoichiometry:

1 mole of AlCl3 produces 1 mole of Al (OH) 3

# moles of Al (OH) 3 = 0.1005 moles

Molar mass Al (OH) 3 = 78 g/mol

Amount of Al (OH) 3 produced = 0.1005*78 = 7.84 g

c) Given mass of NaOH = 10.0 g

Molar mass of NaOH = 40 g/mol

# moles of NaOH = 10/40 = 0.25 moles

Based on the reaction stoichiometry:

3 moles of NaOH produces 1 mole of Al (OH) 3

# moles of Al (OH) 3 produced = 0.25/3 = 0.0833 moles

Molar mass Al (OH) 3 = 78 g/mol

Amount of Al (OH) 3 produced = 0.0833*78 = 6.50 g

d) Since moles of Al (OH) 3 produced from NaOH is less, NaOH will be the limiting reagent. The amount of Al (OH) 3 actually produced is 6.50 g