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Answer:

0.675 g HCl

0.823 g AlCl₃

Explanation:

Step 1: Given data

Mass of Al(OH)₃: 0.500 g

Step 2: Write the balanced equation

3 HCl + Al(OH)₃ ⇒ AlCl₃ + 3 H₂O

Step 3: Calculate the number of grams of HCl that can react with 0.500 g of Al(OH)₃

We will use the following relationships.

  • The molar mass of Al(OH)₃ is 81.03 g/mol.
  • The molar ratio of HCl to Al(OH)₃ is 3:1.
  • The molar mass of HCl is 36.46 g/mol.

[tex]0.500gAl(OH)_3 \times \frac{1molAl(OH)_3}{81.03gAl(OH)_3} \times \frac{3molHCl}{1molAl(OH)_3} \times \frac{36.46gHCl}{1molHCl} = 0.675 g HCl[/tex]

Step 4: Calculate the number of grams of AlCl₃ formed when 0.500 g of Al(OH)₃ reacts

We will use the following relationships.

  • The molar mass of Al(OH)₃ is 81.03 g/mol.
  • The molar ratio of AlCl₃ to Al(OH)₃ is 1:1.
  • The molar mass of HCl is 133.34 g/mol.

[tex]0.500gAl(OH)_3 \times \frac{1molAl(OH)_3}{81.03gAl(OH)_3} \times \frac{1molAlCl_3}{1molAl(OH)_3} \times \frac{133.34gAlCl_3}{1molAlCl_3} = 0.823 g AlCl_3[/tex]

Oseni

The mass of HCl that can react with 0.500 g of Al(OH)3 would be 0.702 g while the mass of AlCl3 that would form when 0.500g Al(OH)3 reacts will be 0.853 g.

From the balanced equation of the reaction:

[tex]Al(OH)_3 + 3 HCl ---> AlCl_3 + 3 H_2O[/tex]

The mole ratio of Al(OH)3 to HCl is 1:3.

mole of 0.5g Al(OH)3 = mass/molar mass

                                 = 5/78

                                    = 0.0064 moles

Thus, equivalent moles of HCl = 0.064 x 3

                                                 = 0.0192 moles

Mass of 0.192 moles HCl = mole x molar mass

                                         = 0.192 x 36.5

                                          = 0.702 g

c) From the balanced equation, the mole ratio of Al(OH)3  to AlCl3 is 1:1.

mole of 0.5g Al(OH)3  = 0.0064 moles

Equivalent mole of AlCl3 = 0.0064 moles

Mass of 0.0064 moles of AlCl3 = 0.0064 x 133.34

                                                = 0.853 g

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886?referrer=searchResults

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