Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 78 mL of water from 29 ∘C to 78 ∘C.

We need 1.1 grams of Mg

Explanation:

Step 1: Data given

Volume of water = 78 mL

Initial temperature = 29 °C

Final temperature = 78 °C

The standard heats of formation

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Step 2: The equation

The heat is produced by the following reaction:

Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)

Step 3: Calculate the mass of Mg needed

Using the standard heats of formation:

−285.8 kJ/mol H2O(l)

−924.54 kJ/mol Mg(OH)2(s)

Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)

−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg

(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required

(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg

We need 1.1 grams of Mg

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