if 15 grams of ethane gas c2h6, reacts with 128 g oxygen un a combustion reaction, how many moles of water can be produced?
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
2C2H6 + 7O2 ==> 4CO2 + 6H2O
mols C2H6 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols ethane to mols H2O
Do the same and convert mols O2 to mols H2O.
It is likely that the two values for mols H2O will not be the same which means one is not correct; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR.
To determine the number of moles of water that can be produced in the combustion reaction between ethane (C2H6) and oxygen (O2), we need to follow these steps:
1. Write the balanced chemical equation for the combustion reaction:
C2H6 + O2 -> CO2 + H2O
2. Find the molar mass of ethane (C2H6) and water (H2O):
- The molar mass of ethane (C2H6):
Molar mass of carbon (C) = 12.01 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Total molar mass of ethane = (2 * molar mass of carbon) + (6 * molar mass of hydrogen) = (2 * 12.01) + (6 * 1.01) = 30.07 g/mol
- The molar mass of water (H2O):
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Total molar mass of water = (2 * molar mass of hydrogen) + (1 * molar mass of oxygen) = (2 * 1.01) + 16.00 = 18.02 g/mol
3. Calculate the number of moles of ethane (C2H6) and oxygen (O2) based on their masses:
- Moles of ethane (C2H6) = Mass of ethane (g) / Molar mass of ethane (g/mol)
= 15 g / 30.07 g/mol
- Moles of oxygen (O2) = Mass of oxygen (g) / Molar mass of oxygen (g/mol)
= 128 g / 32.00 g/mol (assuming O2 gas)
4. Determine the limiting reactant:
To identify the limiting reactant, we need to compare the moles of ethane and oxygen. The reactant that produces fewer moles of water is the limiting reactant.
Using the balanced chemical equation, we can see that 1 mole of ethane produces 3 moles of water.
So, the moles of water produced from ethane = Moles of ethane x (3 moles of water / 1 mole of ethane)
5. Calculate the moles of water:
Moles of water produced = Moles of ethane x (3 moles of water / 1 mole of ethane)
Now, you can plug in the values and calculate the moles of water produced.