Meals-ready-to-eat (MREs) are military meals that can be heated on a flameless heater. The heat is produced by the following reaction: $$\mathrm{Mg}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g)$$ (a) Calculate the standard enthalpy change for this reaction. (b) Calculate the number of grams of Mg needed for this reaction to release enougy energy to increase the temperature of 75 mL of water from 21 to \(79^{\circ} \mathrm{C}\) .

To calculate the standard enthalpy change, ΔH, for the reaction, we use the following formula: \(ΔH = ∑(standard \ enthalpy \ of \ formation \ of \ products) \ -∑(standard \ enthalpy \ of \ formation \ of \ reactants)\) The standard enthalpies of formation for the given species are: Mg (s): 0 kJ/mol (since it is an element in its standard state) \(H_{2}O (l)\): -285.8 kJ/mol \(Mg(OH)_{2} (s)\): -924.42 kJ/mol \(H_{2} (g)\): 0 kJ/mol (since it is an element in its standard state) Now, we'll plug in these values into the equation: \(ΔH = [(1 \times (-924.42)) + (2 \times 0)] - [(1 \times 0) + (2 \times (-285.8))]\)

By computing the above expression, we find: \(\)ΔH = (-924.42) - (-571.6) = -352.82 kJ/mol This is the standard enthalpy change for the reaction.

To find the mass of Mg needed to produce enough heat to increase the temperature of 75 mL of water by the given amount, we will use the following equations: 1) \(q = mcΔT\) (where q represents heat, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature) 2) \(\frac{ΔH}{ΔT} = \frac{mass \ of \ Mg \ used}{molar \ mass \ of \ Mg}\) (where ΔH is the standard enthalpy change, ΔT is the change in temperature, and the molar mass of Mg is 24.305 g/mol) We know that the specific heat capacity of water, c, is 4.18 J/g°C and that the density of water is approximately 1 g/mL. First, let's find the mass of the 75 mL of water: mass of water = (density of water) × (volume of water) = 1 g/mL × 75 mL = 75 g Now, calculate the heat required, q, to increase the temperature from 21°C to 79°C: q = mcΔT = (75 g)(4.18 J/g°C)(79-21) = \(75 \times 4.18 \times 58 \approx 14,433.2 \mathrm{J}\), or 14.4332 kJ Since \(ΔH \approx -352.82 \mathrm{kJ/mol}\) for the reaction, we have about 352.82 kJ of heat released for every mole of Mg reacted. Now we can set up a proportion to find the number of grams of Mg needed to release 14.4332 kJ of heat: \(\frac{1 \ mol \ of \ Mg}{-352.82 \ kJ \ heat} = \frac{x \ grams \ of \ Mg}{14.4332 \ kJ \ heat}\) Solve for x, the mass of Mg needed: \(x = Molar\ mass\ of\ Mg\times\frac{14.4332}{352.82} = \frac{24.305 \times 14.4332}{-352.82} \approx 1.0 \ g\) The mass of Mg required to release sufficient heat to increase the temperature of 75 mL of water from 21°C to 79°C is approximately 1.0 gram.