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

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

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