how many grams of water are produced when 15 grams of magnesium hydroxide are reacted with sulfuric acid

9.26 g of water

Explanation:

We are given;

15 grams of magnesium hydroxide;

Required to calculate the mass of water produced;

We will use the following steps;

Step 1: Write the balanced equation for the reaction

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)

Step 2: Determine the number of moles of Mg (OH) ₂ used

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

Step 3: Calculate the moles of water produced

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

Step 4: Calculate the mass of water produced;

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