Question

Mg(OH)₂ is the active ingredient in many laxatives. It can be made with the following

reaction:
Mg (s) + 2H₂O (1)→ Mg(OH)₂ (s) + H₂ (g)
If a laxative manufacturer has 2.8 x 105 grams of Mg, how many grams of water must be
used to have all of the Mg react to form Mg(OH)₂?

Answer 1

The balanced chemical equation for the reaction is:
Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g)

The molar mass of Mg is 24.31 g/mol.
The molar mass of H2O is 18.015 g/mol.

To calculate the number of moles of Mg present, we divide the mass of Mg by its molar mass:
n(Mg) = 2.8 x 10^5 g / 24.31 g/mol = 11523.6 mol

According to the balanced equation, 1 mole of Mg reacts with 2 moles of H2O to produce 1 mole of Mg(OH)2. Therefore, we need twice the number of moles of H2O as that of Mg:
n(H2O) = 2 x n(Mg) = 2 x 11523.6 mol = 23047.2 mol

Finally, we can calculate the mass of H2O required using the number of moles and its molar mass:
mass(H2O) = n(H2O) x molar mass(H2O)
mass(H2O) = 23047.2 mol x 18.015 g/mol = 415,426.3 g

Therefore, the manufacturer needs 415,426.3 grams of water to react with 2.8 x 10^5 grams of Mg to form Mg(OH)2.