Answer:
120.2 g of aluminum sulfate will be produced
Step-by-step explanation:
The balanced chemical equation for the reaction between aluminum hydroxide and sulfuric acid is:
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
According to the equation, it takes 2 moles of aluminum hydroxide to react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate.
To find out how many grams of aluminum sulfate will be produced, we first need to calculate the number of moles of each reactant.
The molar mass of aluminum hydroxide (Al(OH)3) is:
Al = 26.98 g/mol
O = 15.99 g/mol (3 atoms)
H = 1.01 g/mol (9 atoms)
Total = 78.00 g/mol
Therefore, 54.9 g of aluminum hydroxide is equal to 54.9 g / 78.00 g/mol = 0.7038 moles of aluminum hydroxide.
The molar mass of sulfuric acid (H2SO4) is:
H = 1.01 g/mol (2 atoms)
S = 32.07 g/mol
O = 15.99 g/mol (4 atoms)
Total = 98.08 g/mol
Therefore, 91.9 g of sulfuric acid is equal to 91.9 g / 98.08 g/mol = 0.9361 moles of sulfuric acid.
Since 2 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate, the limiting reactant in this case is aluminum hydroxide, as it is present in the smaller amount.
The number of moles of aluminum sulfate produced can be calculated as:
0.7038 moles Al(OH)3 × (1 mole Al2(SO4)3 / 2 moles Al(OH)3) = 0.3519 moles Al2(SO4)3
The molar mass of aluminum sulfate (Al2(SO4)3) is:
Al = 26.98 g/mol (2 atoms)
S = 32.07 g/mol (3 atoms)
O = 15.99 g/mol (12 atoms)
Total = 342.15 g/mol
Therefore, the mass of aluminum sulfate produced is:
0.3519 moles Al2(SO4)3 × 342.15 g/mol = 120.2 g
Therefore, 120.2 g of aluminum sulfate will be produced.