When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O, how many grams of water are formed?
balance the equation:
2C2H6+5O2>>4CO2 + 6H2O
figure the moles in 20 g ethane, and moles in 60 grams O2.
Notice, that you need 2.5x moles O2 as moles ethane. Do you have that much? If more, then ethane is the limiting reactant. If not, thenO2 is limiting.
so if ethane is limiting, then you get 3 times the moles of water that you have of ethane, if O2 is limiting, you get 6/5 the moles you have of O2.
To find the number of grams of water formed, we need to determine the limiting reactant of the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
First, we need to write the balanced chemical equation for the reaction between C2H6 and O2:
2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O
Now we can calculate the number of moles for each reactant:
Molar mass of C2H6 = 2 * (12.01 g/mol) + 6 * (1.01 g/mol) = 30.07 g/mol
Molar mass of O2 = 2 * (16.00 g/mol) = 32.00 g/mol
Number of moles of C2H6 = 20.0 g / 30.07 g/mol ≈ 0.665 mol
Number of moles of O2 = 60.0 g / 32.00 g/mol ≈ 1.875 mol
Next, we calculate the mole ratio of the reactants based on the balanced equation. For every 2 moles of C2H6, 7 moles of O2 are required to react:
Mole ratio of C2H6 to O2 = 2/7
Since the mole ratio is less than 1, it means that there is an excess of C2H6 and O2 will be the limiting reactant. This means that all the C2H6 will react, and only a certain amount of O2 will react to form products.
Now, we need to calculate the number of moles of water formed. From the balanced equation, we know that for every 6 moles of water formed, 7 moles of O2 will react:
Mole ratio of O2 to H2O = 7/6
Number of moles of H2O = (1.875 mol O2) * (6 mol H2O / 7 mol O2) ≈ 1.607 mol
Finally, we calculate the mass of water formed using the molar mass of water:
Molar mass of H2O = 2 * (1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of water formed = Number of moles of H2O * Molar mass of H2O
= 1.607 mol * 18.02 g/mol ≈ 28.99 g
Therefore, approximately 29.0 grams of water are formed in the given reaction.