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.
[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.
[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]
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
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