How many grams of MgO are produced during an enthalpy change of -96.0 KJ if 2Mg(s) + O2(g) -> 2MgO(s) and delta H= -1204kj
2Mg + O2 ==> 2MgO
2*molar mass MgO x (96/1204) = ? kJ.
Well, to calculate that, we need to use stoichiometry. First, let's convert the given enthalpy change into grams:
Given:
∆H = -96.0 kJ
Molar mass of MgO = 40.31 g/mol
So, the number of moles of MgO produced can be calculated using the equation:
∆H = (n * MW) / ΔT
Where:
n = number of moles
MW = molar mass
ΔT = change in temperature (which is 1K at constant pressure)
Rearranging the equation to solve for n:
n = (∆H * ΔT) / MW
Plugging in the values:
n = (-96.0 kJ * 1K) / 40.31 g/mol
Calculating:
n = -2.38 kJ / 40.31 g/mol
n ≈ -0.059 mol
Since the reaction is 2Mg(s) + O2(g) -> 2MgO(s), the stoichiometric ratio is 2:2. This means that for every 2 moles of MgO, we get 2 moles of Mg.
So, for -0.059 mol of MgO, we get -0.059 mol of Mg. However, we cannot have negative moles, so we'll take the absolute value.
The number of grams of MgO is given by:
grams = n * MW
Plugging the values:
grams = |(-0.059 mol)| * 40.31 g/mol
Calculating:
grams ≈ 2.38 g
So, approximately 2.38 grams of MgO are produced during this enthalpy change.
To determine the number of grams of MgO produced, we need to use the given enthalpy change (∆H) and the stoichiometry of the balanced equation.
The balanced equation is: 2Mg(s) + O2(g) → 2MgO(s)
Given:
∆H = -1204 kJ
We want to find the mass of MgO produced, so we need to use the stoichiometric ratio between MgO and the given enthalpy change.
The molar ratio between MgO and the enthalpy change can be determined by comparing the coefficients in the balanced equation:
2Mg(s) + O2(g) → 2MgO(s)
According to the balanced equation, the molar ratio between MgO and the enthalpy change is 2:1.
Now, we can set up a proportion using the molar ratio:
∆H / 2MgO = -1204 kJ / x grams
Solving for x (grams of MgO):
∆H = -1204 kJ
∆H / 2 = -1204 kJ / 2
∆H / 2 = -602 kJ
Now we can rewrite the proportion:
-602 kJ / 1 = -96.0 kJ / x grams
Cross-multiplying to solve for x:
(-602 kJ) * (x grams) = (-96.0 kJ) * (1)
-602x = -96.0
x = -96.0 / -602
x ≈ 0.1595 grams
Therefore, approximately 0.1595 grams of MgO are produced during an enthalpy change of -96.0 kJ.
To determine the number of grams of MgO produced during the given enthalpy change, we need to utilize stoichiometry and the concept of molar mass.
First, let's convert the enthalpy change of -96.0 kJ to moles. We know that the enthalpy change is given by -1204 kJ for the reaction 2Mg(s) + O2(g) -> 2MgO(s). Therefore, we can set up the following proportion:
(-96.0 kJ) / (-1204 kJ) = x moles / 1
Cross-multiplying and solving for x, we get:
x = (-96.0 kJ * 1 mole) / (-1204 kJ)
x ≈ 0.0797 moles
Next, we need to use the stoichiometry of the balanced equation to calculate the moles of MgO produced. From the equation 2Mg(s) + O2(g) -> 2MgO(s), we can see that 2 moles of MgO are produced for every 2 moles of Mg(s) reacted. Therefore, the moles of MgO produced will be the same as the moles of Mg(s) reacted in this case.
Moles of MgO produced = 0.0797 moles
Finally, we can calculate the mass of MgO produced using the molar mass of MgO. The molar mass of MgO is calculated by adding the atomic masses of Mg and O:
MgO = (24.31 g/mol for Mg) + (16.00 g/mol for O)
MgO = 40.31 g/mol
Mass of MgO produced = moles of MgO produced * molar mass of MgO
Mass of MgO produced = 0.0797 moles * 40.31 g/mol
Mass of MgO produced ≈ 3.20 grams
Therefore, approximately 3.20 grams of MgO are produced during this enthalpy change of -96.0 kJ.