I need help with this question. No links, a real answer. Thank you NaOH + AlCl3 --> Al(OH)3 + NaCI.
How many grams of Al(OH)3 are produced from 3.00 g of AlCl3 with excess of NaOH?

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.

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