WILL AWARD BRAINLIEST!!! A chemist performs a reaction combining 100.1 g of aluminum chloride (AlCl3), and 57.2 g of sodium hydroxide (NaOH). The products are aluminum hydroxide (Al(OH)3) and sodium chloride (NaCl).


a. Please write out the balanced equation for this reaction.


b. Assuming all the aluminum chloride is consumed in this reaction, how many grams of aluminum hydroxide will form? Work must be shown to earn credit.


c. If all the sodium hydroxide is used up, how many grams of aluminum hydroxide can be formed? Work must be shown to earn credit.


d. What is the theoretical yield of this reaction? Which reagent is limiting? Explain your answer.

Respuesta :

a.The balanced reaction between [tex]AlCl_3[/tex] and NaOH is:

[tex]AlCl_3 + 3NaOH[/tex] → [tex]Al(OH)_3 + 3NaCl[/tex] is the balanced reaction.

b.58.55g of aluminum hydroxide will form? Work must be shown to earn credit

c. 6.50 g grams of aluminum hydroxide can be formed.

d.[tex]Al(OH)_3[/tex] produced from NaOH is less, NaOH will be the limiting reagent. The amount of [tex]Al(OH)_3[/tex] actually produced is  4.13 g.

What is a balanced chemical equation?

An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.

a.The balanced reaction between [tex]AlCl_3[/tex] and NaOH is:

[tex]AlCl_3 + 3NaOH[/tex] → [tex]Al(OH)_3 + 3NaCl[/tex] is the balanced reaction.

b) Given mass of [tex]AlCl_3[/tex] = 100.1 g

Molar mass of [tex]AlCl_3[/tex] = 133.34 g/mol

moles of [tex]AlCl_3[/tex] = 100.1 g ÷133.34 = 0.7507 moles

Based on the reaction stoichiometry:

1 mole of [tex]AlCl_3[/tex] produces 1 mole of [tex]Al(OH)_3[/tex]

# moles of [tex]Al(OH)_3[/tex] = 0.7507 moles

Molar mass [tex]Al(OH)_3[/tex] = 78 g/mol

Amount of [tex]Al(OH)_3[/tex] produced = 0.7507 moles x 78 = 58.55g

c) Given mass of NaOH = 57.2 g

Molar mass of NaOH= 40 g/mol

moles of NaOH = 57.2 g ÷ 40 = 1.43 moles

Based on the reaction stoichiometry:

3 moles of NaOH produces 1 mole of [tex]Al(OH)_3[/tex]

moles of [tex]Al(OH)_3[/tex] produced = 1.43÷ 3 = 0.4766 moles

Molar mass [tex]Al(OH)_3[/tex] = 78 g/mol

Amount of [tex]Al(OH)_3[/tex] produced = 0.4766 78 = 6.50 g

d) Since moles of [tex]Al(OH)_3[/tex] produced from NaOH is less, NaOH will be the limiting reagent. The amount of [tex]Al(OH)_3[/tex] actually produced is 6.50 g.

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