For the complete combustion of 1.000 mole of ethane gas at 298K and 1 atm pressure, change in H= -1560 kJ/mol. What will be the heat released when 4.42g of ethane is combusted under these conditions?

2C2H6 + 7O2 ==> 4CO2 + 6H2O

How many mols ethane do you have? That's 4.42g/molar mass ethane = approx 0.15 but you need a more accurate answer.
Then you know you get 1560 kJ for 1 mol so you will get ? for 0.15 mol?

8945

To find the heat released when 4.42g of ethane is combusted, we first need to convert the mass of ethane to moles.

1 mole of ethane (C2H6) has a molar mass of 30.07 g/mol.

To convert grams to moles, we can use the formula:

moles = mass (g) / molar mass (g/mol)

Let's calculate the moles of ethane:

moles = 4.42 g / 30.07 g/mol

moles = 0.147 mol

Now, we can use the given change in enthalpy (ΔH) and the number of moles of ethane to calculate the heat released during combustion.

The change in enthalpy (ΔH) of -1560 kJ/mol corresponds to the combustion of 1 mole of ethane.

To calculate the heat released when 0.147 mol of ethane is combusted, we can use the following proportion:

-1560 kJ/mol = x kJ / 0.147 mol

Solving for x (the heat released):

x = -1560 kJ/mol * 0.147 mol

x = -229.32 kJ

Therefore, the heat released when 4.42 g of ethane is combusted under these conditions is approximately -229.32 kJ.

To find the heat released when 4.42 g of ethane is combusted, we first need to calculate the number of moles of ethane.

The molar mass of ethane (C2H6) is:
2 * atomic mass of carbon + 6 * atomic mass of hydrogen
= 2 * 12.01 g/mol + 6 * 1.01 g/mol
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol

Now, we can calculate the number of moles of ethane:
moles = mass / molar mass
moles = 4.42 g / 30.08 g/mol
moles ≈ 0.1469 mol

Next, we can use the given change in enthalpy (ΔH) for complete combustion of 1.000 mole of ethane to calculate the heat released:
ΔH = -1560 kJ/mol

To find the heat released for 0.1469 mol of ethane, we can use the following proportion:

Heat released (kJ) = (ΔH * moles of ethane) / 1 mole
Heat released (kJ) = (-1560 kJ/mol * 0.1469 mol) / 1 mole

Now, we can substitute the values and calculate the heat released:
Heat released (kJ) ≈ -1560 kJ/mol * 0.1469 mol
Heat released (kJ) ≈ -228.764 kJ

Therefore, the heat released when 4.42 g of ethane is combusted under these conditions is approximately -228.764 kJ.