For the combustion of acetylene:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) ΔH = -2511 kJ
What is the enthalpy change when 2.80 g of C2H2 (g) reacts with excess oxygen.
Well, if we have 2.80 g of C2H2 (g), let's start by converting that to moles.
The molar mass of C2H2 is 26.04 g/mol, so 2.80 g of C2H2 would be:
2.80 g / 26.04 g/mol = 0.1074 mol
Since the balanced equation tells us that the coefficient of C2H2 is 2, we can see that for every 2 moles of C2H2, the enthalpy change is -2511 kJ.
Therefore, we can set up a proportion:
2 mol C2H2 / -2511 kJ = 0.1074 mol C2H2 / x kJ
Solving for x, we find:
x = (0.1074 mol C2H2 * -2511 kJ) / 2 mol C2H2
x ≈ -135.6 kJ
So, the enthalpy change when 2.80 g of C2H2 (g) reacts with excess oxygen is approximately -135.6 kJ. Keep in mind, though, that my calculations might be more accurate than my jokes!
To find the enthalpy change when 2.80 g of C2H2 (g) reacts with excess oxygen, we need to use the given enthalpy change value.
1. First, we need to determine the number of moles of C2H2(g) in 2.80 g.
To do this, we will use the molar mass of C2H2, which is:
C: 12.01 g/mol x 2 = 24.02 g/mol
H: 1.008 g/mol x 2 = 2.016 g/mol
Total molar mass of C2H2 = 24.02 g/mol + 2.016 g/mol = 26.036 g/mol
Now, we can calculate the number of moles:
Moles of C2H2 = Mass / Molar mass
Moles of C2H2 = 2.80 g / 26.036 g/mol
2. Next, we need to determine the number of moles of O2 required for the complete combustion of C2H2.
From the balanced chemical equation, we can see that 2 moles of C2H2 react with 5 moles of O2.
3. Calculate the number of moles of O2 required:
Moles of O2 = (Moles of C2H2) x (5 moles of O2 / 2 moles of C2H2)
4. Finally, we can find the enthalpy change using the equation:
Enthalpy change = (Moles of O2) x (ΔH)
Substitute the values and calculate the enthalpy change.
To calculate the enthalpy change when 2.80 g of C2H2 (g) reacts, we need to use the balanced equation and apply stoichiometry.
Step 1: Convert the mass of C2H2 (g) to moles
To do this, we'll use the molar mass of C2H2. The molar mass of C is 12.01 g/mol, and the molar mass of H is 1.01 g/mol. Since there are two carbons and two hydrogens in C2H2, the molar mass of C2H2 is:
(2 * 12.01 g/mol) + (2 * 1.01 g/mol) = 26.04 g/mol
Now we can calculate the number of moles of C2H2:
2.80 g / 26.04 g/mol = 0.1074 mol
Step 2: Use the balanced equation to determine the mole ratio
According to the balanced equation, the stoichiometric ratio of C2H2 to ΔH is 2:2511 kJ. That means for every 2 moles of C2H2, the enthalpy change is -2511 kJ.
Step 3: Calculate the enthalpy change
We can use the mole ratio obtained in step 2 to find the enthalpy change when 0.1074 mol of C2H2 reacts.
0.1074 mol C2H2 * (-2511 kJ / 2 mol C2H2) = -134.6 kJ
Therefore, the enthalpy change when 2.80 g of C2H2 (g) reacts with excess oxygen is approximately -134.6 kJ.