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