The combustion of how many moles of ethane would be required to heat 838 grams of water from 25.0 to 93.0 degrees C? Assume liquid water is formed during the combustion.
To determine the number of moles of ethane required for combustion, we need to follow these steps:
Step 1: Calculate the amount of heat required to raise the temperature of water from 25.0 to 93.0 degrees Celsius.
The equation to calculate the heat is: Q = m × C × ΔT, where Q is the heat, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Given:
- Mass of water (m) = 838 grams
- Change in temperature (ΔT) = 93.0°C - 25.0°C = 68.0°C
The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, substituting these values into the equation:
Q = 838 g × 4.18 J/g°C × 68.0°C
Step 2: Calculate the amount of heat released by the combustion of ethane.
The combustion reaction of ethane (C2H6) can be written as:
C2H6 + 7/2 O2 → 2CO2 + 3H2O
From the balanced equation, we can see that for every mole of ethane combusted, 3 moles of water are formed.
Step 3: Convert the heat released by the combustion into moles of ethane.
Using the molar mass of ethane (30.07 g/mol) and the heat of formation of water (-285.8 kJ/mol), we can calculate the moles of ethane required to produce the given amount of heat.
To summarize, the steps are:
1. Calculate the amount of heat required to heat the water.
2. Calculate the amount of heat released by the combustion of ethane.
3. Convert the heat released into moles of ethane.
Shall we proceed with the calculation?
To solve this problem, we need to follow a few steps:
Step 1: Calculate the amount of heat required to raise the temperature of water. We can use the equation:
q = m * C * ΔT
where:
- q is the amount of heat in joules
- m is the mass of water in grams (838 grams)
- C is the specific heat capacity of water (4.184 J/g°C)
- ΔT is the change in temperature (93.0 - 25.0 = 68.0 °C)
Using these values, we can calculate the amount of heat required:
q = 838 g * 4.184 J/g°C * 68.0 °C = 234,488.352 J
Step 2: Convert the amount of heat to moles of ethane.
To do this, we need to use the balanced chemical equation for the combustion of ethane (C2H6):
C2H6 + 7/2 O2 → 2 CO2 + 3 H2O
From the equation, we can see that for every 3 moles of water formed, 1 mole of ethane is required.
Step 3: Calculate the moles of water formed during combustion.
Since the balanced equation states that 3 moles of water are formed for every mole of ethane, we can convert the mass of water to moles:
moles of water = mass of water / molar mass of water
The molar mass of water (H2O) is 18.015 g/mol.
moles of water = 838 g / 18.015 g/mol = 46.524 mol
Step 4: Calculate the moles of ethane required.
Since the ratio between moles of ethane and moles of water is 1:3, we can calculate the moles of ethane:
moles of ethane = moles of water / 3
moles of ethane = 46.524 mol / 3 = 15.508 mol
Therefore, the combustion of approximately 15.508 moles of ethane would be required to heat 838 grams of water from 25.0 to 93.0 degrees Celsius.
q is the amount of heat you will need to heat the 838 g water.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Do you have the heat of combustion? If not you will need to calculate it.
2C2H6(g) + 7O2(g) ==> 4CO2(g) + 6H2O(l)
dHrxn = (n*dHf products) - (n*dHf reactants)
dHrxn is the heat of combustion.
dHrxn/2 x mol = q
Solve for mol. Remember to keep the units for dHrxn and q the same.