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
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.
Answer:
11.43g of aluminum hydroxide!
Explanation: