What mass of water is produced when 0.1 moles of C2H6 is reacted with 0.1 moles of
O2?
2C2H6 + 7O2 => 4CO2 + 6H2O
This is a limiting reagent problem; you know that because amounts are given for BOTH reactants. Therefore, you must identify the limiting reagent.
How much H2O can be produced by 0.1 mol C2H6. That is
0.1 mols C2H6 x (6 mols H2O/2 mols C2H6) = ?
How much H2O can be produced by 0.1 mol O2? That is
01 mol O2 x (6 mols H2O/7 mols O2) = ?
The mols of water will be the SMALLER of the two.
Then convert to grams. grams = mols H2O x molar mass H2O = ?
To determine the mass of water produced, we first need to determine the balanced chemical equation for the reaction between C2H6 (ethane) and O2 (oxygen).
The balanced equation is as follows:
C2H6 + O2 --> H2O + CO2
From the balanced equation, we can see that for every 1 mole of C2H6, 3 moles of H2O are produced. Therefore, we can use the ratio of moles to determine the moles of water produced.
Given that we have 0.1 moles of C2H6, we can multiply it by the ratio of moles to find the moles of water produced:
0.1 moles C2H6 x (3 moles H2O / 1 mole C2H6) = 0.3 moles H2O
Now, we need to calculate the mass of water produced using the molar mass of water (H2O), which is approximately 18.015 g/mol.
Mass of water = Moles of water x Molar mass of water
Mass of water = 0.3 moles H2O x 18.015 g/mol = 5.4055 g
Therefore, the mass of water produced when 0.1 moles of C2H6 is reacted with 0.1 moles of O2 is approximately 5.4055 grams.