In the reaction,C2H4(g) + 3O2(g) =2H2O(g) + 2CO2(g),what is the complete combustion of 12.5cm3 of ethene

what is the complete combustion of 12.5cm3 of ethene >> makes no sense to me.

To determine the complete combustion of 12.5 cm3 of ethene (C2H4), we need to calculate the amount of ethene in moles and use the stoichiometric coefficients from the balanced equation to find the amounts of water (H2O) and carbon dioxide (CO2) produced.

Step 1: Calculate the moles of ethene (C2H4)
To calculate the moles of ethene, we need to know the molar volume of ethene at a given temperature and pressure, which is usually given in the question or can be looked up in a reference table. Let's assume that the molar volume of ethene at standard temperature and pressure (STP) is 22.4 L/mol.

moles of ethene = (volume of ethene in liters) / (molar volume of ethene)

Since the given volume is in cm3, we need to convert it to liters:
volume of ethene in liters = 12.5 cm3 * (1 L/1000 cm3) = 0.0125 L

moles of ethene = 0.0125 L / 22.4 L/mol = 0.000558 moles

Step 2: Use the stoichiometric coefficients to calculate the moles of water and carbon dioxide produced
From the balanced equation, we know that the ratio of moles of ethene to moles of water is 1:2 and the ratio of moles of ethene to moles of carbon dioxide is 1:2.

moles of water = 2 * moles of ethene = 2 * 0.000558 moles = 0.001116 moles
moles of carbon dioxide = 2 * moles of ethene = 2 * 0.000558 moles = 0.001116 moles

Step 3: Convert moles of water and carbon dioxide to volumes (if needed)
If the question asks for the volume of water and carbon dioxide produced, you can use the molar volume of water (18 mL/mol) and carbon dioxide (22.4 L/mol) at STP to calculate the volumes.

volume of water = moles of water * molar volume of water = 0.001116 moles * 18 L/mol = 0.02009 L or 20.09 mL
volume of carbon dioxide = moles of carbon dioxide * molar volume of carbon dioxide = 0.001116 moles * 22.4 L/mol = 0.02498 L or 24.98 mL

Therefore, the complete combustion of 12.5 cm3 of ethene yields approximately 20.09 mL of water and 24.98 mL of carbon dioxide.

To find the complete combustion of 12.5 cm3 of ethene (C2H4), we need to determine the ratio of reactants and products in the balanced equation.

1. First, write and balance the equation:
C2H4(g) + 3O2(g) → 2H2O(g) + 2CO2(g)

2. Determine the molar ratio between ethene (C2H4) and carbon dioxide (CO2) in the balanced equation:
From the balanced equation, we can see that 1 mol of C2H4 reacts to produce 2 mol of CO2.

3. Convert the given volume of ethene (12.5 cm3) to moles:
To convert the volume of gas to moles, we need to use the ideal gas equation. However, since the pressure and temperature are not given, we cannot directly convert the volume to moles.

4. Assuming standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere pressure, we can use the molar volume of an ideal gas, which is about 22.4 L/mol or 22.4 dm3/mol.

To convert from cm3 to dm3, divide by 1000:
12.5 cm3 = 12.5/1000 dm3 = 0.0125 dm3

5. Now we can calculate the number of moles of ethene (C2H4):
Volume (dm3) = n (moles) x molar volume (dm3/mol)
0.0125 dm3 = n (moles) x 22.4 dm3/mol

Rearranging the equation:
n (moles) = 0.0125 dm3 / 22.4 dm3/mol

Calculating the moles of ethene:
n (moles) = 0.0005589 moles

6. Using the mole ratio from the balanced equation, we can determine the number of moles of carbon dioxide (CO2) produced:
For every 1 mole of C2H4, we get 2 mol of CO2.
So, for 0.0005589 moles of C2H4, we get:
(0.0005589 moles C2H4) x (2 moles CO2/1 mole C2H4) = 0.00112 moles CO2

7. Finally, convert moles of carbon dioxide (CO2) to volume:
Using the molar volume of an ideal gas at STP, we can calculate the volume:
Volume (dm3) = n (moles) x molar volume (dm3/mol)
Volume (dm3) = 0.00112 moles x 22.4 dm3/mol

Calculating the volume of carbon dioxide:
Volume (dm3) = 0.025088 dm3

Therefore, the complete combustion of 12.5 cm3 of ethene will produce 0.0251 dm3 (or 25.1 cm3) of carbon dioxide.