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

9.26 g of water

We are given;

Required to calculate the mass of water produced;

We will use the following steps;

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)

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

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

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

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