When the equation for combustion for ethane is balanced using integer coefficients, the ΔH for the reaction = -2834 kJ. How many grams of ethane must be burned in order to heat 277.3 grams of water from 54.0°C to the boiling point and then boil all of it into the gas phase at 100.0°C? The ΔHvap of water = 40.8 kJ/mol and the specific heat of liquid water (SH2O) = 4.184 J/gK.

2C2H6 + 7O2 --> 4CO2 + 6H2O + 2834 kJ

How many kJ does it take to do the water thing.
q1 = heat needed to heat H2O from 54 C to 100.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = heat needed to boil away the water.
q2 = mass H2O x heat vaporiation.
Total heat needed is q1 + q2 and change to kJ.
2*molar mass C2H6 x (kJ needed/2834 kJ) = grams C2H6 needed.
Post your work if you get stuck.

To find the grams of ethane required, we need to calculate the heat energy required to raise the temperature of the water from 54.0°C to the boiling point, and then the heat energy required to vaporize the water.

First, let's calculate the heat energy required to raise the temperature of the water. We'll use the equation:

q1 = m * SH2O * ΔT

Where:
q1 = heat energy required
m = mass of water
SH2O = specific heat of water
ΔT = change in temperature

q1 = 277.3 g * 4.184 J/gK * (100.0°C - 54.0°C)
q1 = 277.3 g * 4.184 J/gK * 46.0°C
q1 = 54366.2864 J

Next, let's calculate the heat energy required to vaporize the water. We'll use the equation:

q2 = ΔHvap * n

Where:
q2 = heat energy required
ΔHvap = enthalpy of vaporization of water
n = moles of water

First, we need to calculate the moles of water using the molar mass of water (18.015 g/mol):

moles of water = mass of water / molar mass of water
moles of water = 277.3 g / 18.015 g/mol
moles of water = 15.408 mol

q2 = 40.8 kJ/mol * 15.408 mol
q2 = 628.1472 kJ

Now, let's calculate the total heat energy required:

total heat energy = q1 + q2
total heat energy = 54366.2864 J + 628.1472 kJ
total heat energy = 54366.2864 J + 628147.2 J
total heat energy = 682513.4864 J

Finally, let's calculate the grams of ethane required using the given ΔH for the combustion of ethane:

ΔH = -2834 kJ = -2834000 J

The equation for the combustion of ethane is:

C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O

The ΔH for the reaction is -2834000 J. Using the coefficients, we can calculate the heat energy released per mole of ethane:

Heat energy per mole of ethane = ΔH / coefficients of ethane
Heat energy per mole of ethane = -2834000 J / 1 mol
Heat energy per mole of ethane = -2834000 J/mol

To calculate the grams of ethane required, we can use the equation:

mass of ethane = total heat energy / heat energy per mole of ethane

mass of ethane = 682513.4864 J / -2834000 J/mol
mass of ethane ≈ -0.241 kg

Since mass cannot be negative, we can conclude that we need approximately 0.241 kg or 241 grams of ethane.

To solve this problem, we need to calculate the amount of heat required to heat the water from 54.0°C to its boiling point and then boil it into the gas phase.

First, let's calculate the heat (q1) required to heat the water from 54.0°C to its boiling point. We can use the equation:

q1 = m * SH2O * ΔT

Where:
m = mass of water = 277.3 grams
SH2O = specific heat of liquid water = 4.184 J/gK
ΔT = change in temperature = (boiling point - initial temperature)

The boiling point of water is 100.0°C, and the initial temperature is 54.0°C.

q1 = 277.3 grams * 4.184 J/gK * (100.0°C - 54.0°C)

Next, let's calculate the heat (q2) required to boil the water into the gas phase. We can use the equation:

q2 = n * ΔHvap

Where:
n = moles of water
ΔHvap = enthalpy of vaporization of water = 40.8 kJ/mol

To calculate the moles of water, we'll use the molar mass of water (H2O) which is approximately 18 g/mol.

moles of water = mass of water / molar mass of water

Finally, we can calculate the total heat (q) required by adding q1 and q2:

q = q1 + q2

Now, let's plug in the values and calculate the result.