Answer :
Final answer:
When 78.0g of aluminum hydroxide reacts with 49.0g sulfuric acid, 54.0g of water is produced.
Explanation:
First, we need to convert the masses of aluminum hydroxide and sulfuric acid to moles. The moles of Al(OH)3 can be obtained by dividing the mass given (78.0g) by the molar mass of Al(OH)3 (78.0 g/mol), resulting in 1 mole. The moles of H2SO4 can be obtained by dividing the mass given (49.0g) by the molar mass of H2SO4 (98.1 g/mol), which gives 0.499 moles.
Next, we can use the balanced chemical equation to determine the mole ratio of water to Al(OH)3. The balanced equation is:
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
From the equation, we can see that 2 moles of Al(OH)3 produce 6 moles of H2O. Since we have 1 mole of Al(OH)3, that would produce 3 moles of H2O. Finally, we can convert moles of water to grams by multiplying the mole ratio by the molar mass of water (18.0 g/mol), resulting in 54.0g of water.
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Final answer:
In the reaction of aluminum hydroxide with sulfuric acid, aluminum sulfate and water are produced. With the provided amounts of reactants, approximately 54g of water is created.
Explanation:
The subject of the question regards the reaction of aluminum hydroxide, Al(OH)3, with sulfuric acid, H2SO4. In this reaction, aluminum sulfate and water are produced. With the balanced chemical equation:
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
This informs us that one mol of Al(OH)3 will produce 3 mol of water (H2O). Given 78.0g of Al(OH)3, which is equivalent to 1 mol, this would react to form 3 mol of water, or approximately 54 g (since the molar mass of water is approximately 18 g/mol).
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