Answer :
The reaction between 78.0 g of aluminum hydroxide (Al(OH)3) and 49.0 g of sulfuric acid (H2SO4) produces water (H2O).
To determine the mass of water produced, we need to calculate the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. Calculate the number of moles for each reactant:
- Moles of Al(OH)3 = mass / molar mass
Moles of Al(OH)3 = 78.0 g / 78.0 g/mol = 1.00 mol
- Moles of H2SO4 = mass / molar mass
Moles of H2SO4 = 49.0 g / 98.1 g/mol = 0.500 mol
2. Use the stoichiometry of the balanced equation to determine the moles of water produced:
The balanced equation for the reaction 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 react with 3 moles of H2SO4 to produce 6 moles of water.
So, the ratio of moles of H2SO4 to moles of water is 3:6 or 1:2.
Since we have 0.500 mol of H2SO4, we can calculate the moles of water produced:
Moles of water = 0.500 mol × (2 mol water / 1 mol H2SO4) = 1.00 mol
3. Calculate the mass of water produced:
Mass of water = moles of water × molar mass of water
Mass of water = 1.00 mol × 18.0 g/mol = 18.0 g
Therefore, when 78.0 g of aluminum hydroxide reacts with 49.0 g of sulfuric acid, 18.0 g of water is produced.
learn more about aluminum hydroxide
https://brainly.com/question/28272266
#SPJ11
