Answer:
1.75 g of Al(OH)₃.
Explanation:
We'll begin by writing the balance equation for the reaction. This is given below:
3NaOH + AlCl₃ —> Al(OH)₃ + 3NaCI
Next, we shall determine the mass of AlCl₃ that reacted and the mass of Al(OH)₃ produced from the balanced equation equation. This can be obtained as follow:
Molar mass of AlCl₃ = 27 + (3×35.5)
= 27 + 106.5
= 133.5 g/mol
Mass of AlCl₃ from the balanced equation = 1 × 133.5 g = 133.5 g
Molar mass of Al(OH)₃ = 27 + 3(16 + 1)
= 27 + 3(17)
= 27 + 51
= 78 g/mol
Mass of Al(OH)₃ from the balanced equation = 1 × 78 = 78 g
SUMMARY:
From the balanced equation above,
133.5 g of AlCl₃ reacted to produce 78 g of Al(OH)₃.
Finally, we shall determine the mass of Al(OH)₃ produced by the reaction of 3 g of AlCl₃. This can be obtained as follow:
From the balanced equation above,
133.5 g of AlCl₃ reacted to produce 78 g of Al(OH)₃.
Therefore, 3 g of AlCl₃ will react to produce = (3 × 78)/133.5 = 1.75 g of Al(OH)₃.
Thus, 1.75 g of Al(OH)₃ were obtained from the reaction.
