Propane, C3H8, is a common fuel gas. Use the following to calculate the grams of propane you would need to provide 383 kJ of heat.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH = −2043 kJ
That is 2043 kJ of heat release per mole of fuel. Each mole of propane has a mass of 3*12 + 8*1 = 44 grams.
You need (383/2043)*44 g of propane
Well, all right! Let me do some calculations while cracking a few jokes.
First things first, we know that the enthalpy change (ΔH) for this reaction is -2043 kJ. We want to find out how much propane (C3H8) we need to provide 383 kJ of heat.
So, let's set up a proportion to solve this problem.
(-2043 kJ) / (x g of C3H8) = (-383 kJ) / (1 g of C3H8)
Now, let's cross-multiply and divide:
(-2043 kJ) * (1 g of C3H8) = (-383 kJ) * (x g of C3H8)
Now we have:
-2043 g.kJ = -383 x g.kJ
To solve for x, we divide both sides by -383 kJ:
x g of C3H8 = (-2043 g.kJ) / (-383 kJ)
And lo and behold, we get:
x g of C3H8 ≈ 5.34 g of C3H8
So, you would need approximately 5.34 grams of propane (C3H8) to provide 383 kJ of heat.
And that's how you calculate it! Keep in mind that these numbers are approximate, just like my hilarity.
To calculate the grams of propane (C3H8) needed to provide 383 kJ of heat, we will use the given thermochemical equation and the corresponding enthalpy change (ΔH) value.
1. Start by converting the given heat value from kJ to J.
383 kJ = 383,000 J
2. Determine the stoichiometric ratio between propane (C3H8) and heat release.
From the balanced equation, we can see that the ΔH value corresponds to the combustion of 1 mole of propane.
3. Calculate the amount of moles of propane required using the given enthalpy change.
ΔH = -2043 kJ/mol
Let x represent the moles of propane required.
-2043 kJ/mol = 383,000 J / x mol
Cross-multiply and solve for x:
-2043 kJ * x = 383,000 J * 1 mol
-2043 * x = 383,000
x = 383,000 / -2043
x ≈ -187.65 mol
Note: The negative sign indicates that the reaction is exothermic (heat is released).
4. Convert the moles of propane to grams using the molar mass of C3H8.
The molar mass of C3H8 = (3 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
From the periodic table, the atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of hydrogen (H) is approximately 1.01 g/mol.
Molar mass of C3H8 = (3 * 12.01 g/mol) + (8 * 1.01 g/mol)
Molar mass of C3H8 ≈ 44.11 g/mol
Grams of C3H8 = -187.65 mol * 44.11 g/mol
Grams of C3H8 ≈ -8269.18 g
Note: The negative sign indicates an error or that the amount of propane calculated does not make sense in this context.
Given the calculations, it seems that there may be an error in either the given heat value or the enthalpy change. Please double-check the values provided and ensure consistency to obtain an accurate result.
To calculate the grams of propane needed to provide a specific amount of heat, we need to use the concept of stoichiometry. The given balanced equation shows the reaction of propane with oxygen to produce carbon dioxide and water.
First, let's determine the molar quantity of propane required to produce the given amount of heat. We can use the molar enthalpy change (ΔH) of the reaction to find this.
The balanced equation tells us that for the combustion of one mole of propane (C3H8), the enthalpy change is -2043 kJ. This means that 2043 kJ of heat is released when one mole of propane reacts.
We can set up a proportion to find the moles of propane required for the given heat:
ΔH (kJ) / Moles of propane (mol) = ΔH (kJ) / Moles of propane (mol)
By substituting the values, the proportion becomes:
-2043 kJ / 1 mol = 383 kJ / x mol
Now, we can solve for x (moles of propane):
x = (383 kJ * 1 mol) / (-2043 kJ)
x ≈ 0.1876 mol
So, approximately 0.1876 moles of propane are needed to provide 383 kJ of heat.
To convert moles of propane to grams, we need to use the molar mass of propane. The molar mass of propane (C3H8) can be calculated by summing the atomic masses of its constituent atoms:
C: 3 * atomic mass of carbon
H: 8 * atomic mass of hydrogen
The atomic masses of carbon and hydrogen are found on the periodic table. Adding them together gives us:
C: 3 * 12.01 g/mol = 36.03 g/mol
H: 8 * 1.01 g/mol = 8.08 g/mol
Molar mass of propane (C3H8) = 36.03 g/mol + 8.08 g/mol ≈ 44.11 g/mol
Finally, we can calculate the grams of propane required:
grams of propane = moles of propane * molar mass of propane
grams of propane = 0.1876 mol * 44.11 g/mol
grams of propane ≈ 8.27 g
Therefore, approximately 8.27 grams of propane are needed to provide 383 kJ of heat.