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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