<h3>
Answer:</h3>
9.26 g of water
<h3>
Explanation:</h3>
We are given;
- 15 grams of magnesium hydroxide;
Required to calculate the mass of water produced;
We will use the following steps;
<h3>Step 1: Write the balanced equation for the reaction</h3>
Magnesium hydroxide reacts with sulfuric acid to produce water and magnesium sulfate;
Therefore, the equation of the reaction is;
Mg(OH)₂(aq) + H₂SO₄(aq) → MgSO₄(aq) + 2H₂O(l)
<h3>Step 2: Determine the number of moles of Mg(OH)₂ used </h3>
Number of moles = Mass ÷ Molar mass
Molar mass of Mg(OH)₂ =58.320 g/mol
Therefore;
Moles of Mg(OH)₂ = 15 g ÷ 58.320 g/mol
= 0.257 moles
<h3>Step 3: Calculate the moles of water produced</h3>
From the equation 1 mole of Mg(OH)₂ reacts to produce 2 moles of water;
Therefore; Moles of water = moles of Mg(OH)₂ × 2
Moles of water = 0.257 moles × 2
= 0.514 moles
<h3>Step 4: Calculate the mass of water produced;</h3>
Mass = Moles × Molar mass
Molar mass of water = 18.02 g/mol
Therefore;
Mass of water = 0.514 moles × 18.02 g/mol
= 9.26 g
Therefore, 9.26 g of water will be produced
