I got through the first part easily, got -352.9kJ
Having trouble with the second part
"The meals-ready-to-eat (MREs) in the military can be heated on a flameless heater. The source of energy in the heater is shown below.
Mg(s) + 2 H2O(l) Mg(OH)2(s) + H2(g)
Calculate the enthalpy change under standard conditions, in joules, for this reaction."
The second part of this problem is....
How many grams of Mg are needed to raise the temperature of 25.0ml of water from 25.0 C to 85.0 C
Energy needed is q.
q = mass H2O x specific heat water x (Tfinal-Tinitial). Then
g Mg = 24.3 x (q kJ/352.9 kJ)
Check my thinking.
To solve the second part of the problem, we need to use the equation:
q = mcΔT
where:
q is the heat energy gained or lost by the water
m is the mass of the water
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature, which is (85.0 - 25.0) = 60.0°C in this case.
First, let's find the mass of water in grams using the density of water:
density = mass/volume
mass = density * volume
The density of water is approximately 1 g/mL, thus:
mass = (1 g/mL) * (25.0 mL) = 25.0 g
Now, substituting the values into the equation:
q = (25.0 g) * (4.18 J/g°C) * (60.0°C)
q = 6270 J
Therefore, 6270 J of heat energy is needed to raise the temperature of 25.0 mL of water from 25.0°C to 85.0°C.
However, we need to determine the mass of magnesium (Mg) needed to generate this amount of energy.
To calculate the mass of Mg, we can use the equation:
Hess's Law: ΔH = q / n
where:
ΔH is the enthalpy change for the reaction (in this case, what we need to find)
q is the heat energy gained or lost (6270 J)
n is the number of moles of Mg consumed or produced, which can be calculated using the balanced equation:
Mg(s) + 2 H2O(l) -> Mg(OH)2(s) + H2(g)
From the balanced equation, we can see that for every 1 mole of Mg, 2 moles of H2O are consumed. Therefore, the moles of Mg is:
n = 2 * (moles of H2O)
To determine the moles of H2O, we can use the molar volume of water at standard temperature and pressure (STP), which is 22.4 L/mol.
mL to L conversion:
25.0 mL = 25.0 / 1000 = 0.025 L
moles of H2O = (0.025 L) / (22.4 L/mol) = 0.00112 mol
Since the stoichiometry ratio is 2 moles of H2O to 1 mole of Mg, we have:
n = 2 * (0.00112 mol) = 0.00224 mol
Now, we can calculate the enthalpy change (ΔH):
ΔH = q / n = 6270 J / 0.00224 mol = 2,795,982 J/mol
Finally, to determine the mass of Mg needed, we need to use the molar mass of Mg, which is approximately 24.31 g/mol:
mass of Mg = moles of Mg * molar mass of Mg
mass of Mg = 0.00224 mol * 24.31 g/mol = 0.0545 g
Therefore, approximately 0.0545 grams of Mg are needed to raise the temperature of 25.0 mL of water from 25.0°C to 85.0°C.