Chemistry, asked by sfsfsdfs, 4 months ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

The table shows the calculated amounts of reactants and products when the reaction was conducted in a laboratory.


Initial Mass and Yield
Sulfuric Acid Aluminum Hydroxide
Initial Amount of Reactant 35 g 30 g
Theoretical Yield of Water from Reactant 12.85 g 31.77 g


What is the approximate amount of the leftover reactant?
8.2 g of sulfuric acid
9.8 g of sulfuric acid
11.43 g of aluminum hydroxide
13.76 g of aluminum hydroxide

Answers

Answered by caffeinated
5

Answer: The approximate mass of the leftover reactant (Aluminium Hydroxide) is 11.43 g.

  • Molar mass of Al(OH)3 = 78 g
  • Molar mass of H2SO4 = 98 g

Given:

Mass of Al(OH)3 =30g

Mass of H2SO4 = 35g

  • Number of moles of Al(OH)3 = 30/78 =0.385
  • Number of moles of H2SO4 = 35/98 =0.357

According to the given reaction,

  2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

  2 moles of Al(OH)3 = 3 moles of H2SO4

  1 moles of Al(OH)3 = 3/2 moles of H2SO4

  0.385 moles of Al(OH)3 = (3/2) x 0.385 moles of H2SO4

                                           = 0.578 moles of H2SO4

But, there is only 0.357 moles of H2SO4 present.

Therefore, H2SO4 is the limiting reagent.

As 0.357 moles of H2SO4 reacts with = 2/3 x 0.357 moles of Al(OH)3

                                                               = 0.238 moles of Al(OH)3

Therefore, leftover Al(OH)3 = 0.385 - 0.238 moles

                                               =  0.147 moles

1 mole of Al(OH)3 = 78 g

0.147 moles = 78 x 0.147 g

                    =    11.43 g

Thus, 11.43 g of Al(OH)3 will be leftover.

Answered by jd213333333
1

Answer:

11.43g of aluminum hydroxide!

Explanation:

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