Determine the number of grams of hcl that can react with 0.750 g of Al(OH)3 according to the following reaction Al(OH)₃(s)+3HCl(aq)→AlCl₃(aq)+3H₂O( aq)

The mass of HCl needed to react with 0.750 g of Al(OH)₃ is 1.05 g.

To solve this question, we'll begin by calculating the masses of Al(OH)₃ and HCl that reacted from the balanced equation. This can be obtained as follow:

Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(aq)

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

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl from the balanced equation = 3 × 36.5 = 109.5 g

Thus,

From the balanced equation above,

78 g of Al(OH)₃ reacted with 109.5 g of HCl

Finally, we shall determine the mass of HCl needed to react with 0.750 g of Al(OH)₃. This can be obtained as follow:

From the balanced equation above,

78 g of Al(OH)₃ reacted with 109.5 g of HCl

Therefore,

0.750 g of Al(OH)₃ will react with = = 1.05 g of HCl.

Thus, we can conclude that the mass of HCl needed to react with 0.750 g of Al(OH)₃ is 1.05 g

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1040 g HCl

Explanation:

Al(OH)₃ + 3HCl => AlCl₃ + 3H₂O

moles Al(OH)₃ = 750g/78g·mol⁻¹ = 9.62 mol

moles HCl used = 3 x moles Al(OH)₃ consumed = 3(9.62)mol HCl = 1038.46 g. ≈ 1040 g HCl (3 sig.figs.)

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tony (lordcaos066)

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