When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O, how many grams of water are formed?
1. Write the balanced chemical equation for the reaction:
2C2H6 + 7O2 = 4CO2 + 6H2O
2. Determine moles C2H6 and O2.
moles C2H6 = 20.0 grams C2H6 x (1 mole C2H6/30.0692 grams C2H6) = 0.665 moles C2H6
moles O2 = 60.0 grams O2 x (1 mole O2/32 grams O2) = 1.875 moles O2
3. Determine limiting reagent. The limiting reagent is the reactant producing the smaller amount of product.
Is it C2H6?
Note: 2 moles C2H6 produce 6 moles H2O
0.665 moles C2H6 x (6 moles H2O/2 moles C2H6) = 1.995 moles H2O
Is it O2?
Note: 7 moles O2 produce 6 moles H2O
1.875 moles O2 x (6 moles H2O/7 moles O2) = 1.607 moles H2O
Therefore, the LR is O2 since it produces the least amount of product.
3. Convert moles H2O to grams H2O.
1.607 mole H2O x (18.02 grams H2O/1 mole H2O) = 28.9 grams H2O
It's 28.9 g.
You're right. You don't have an error in this last post. The limiting reagent is O2 (not H2O) and the 28.9 g H2O is correct.
oh nvmd yes the LR is O2
WHATS THE GRAM OF H2O THEN?
To find the mass of water formed in this chemical reaction, we need to first balance the equation and then use stoichiometry to calculate the amount of water produced.
The balanced equation for the reaction is:
C2H6 + O2 -> CO2 + H2O
In this equation, the ratio of C2H6 to H2O is 1:3. This means that for every 1 mole of C2H6, 3 moles of H2O will be produced.
To calculate the moles of C2H6, we need to convert the given mass of C2H6 to moles. We can use the molar mass of C2H6 (Ethane) to do this:
molar mass of C2H6 = (2 * atomic mass of C) + (6 * atomic mass of H)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 30.07 g/mol
moles of C2H6 = mass of C2H6 / molar mass of C2H6
= 20.0 g / 30.07 g/mol
= 0.665 mol
Now, using the stoichiometric ratio from the balanced equation, we can calculate the moles of H2O formed:
moles of H2O = moles of C2H6 * (3 moles of H2O / 1 mole of C2H6)
= 0.665 mol * (3)
= 1.995 mol
Finally, we can convert the moles of H2O to grams by multiplying by the molar mass of water (H2O):
molar mass of H2O = (2 * atomic mass of H) + (1 * atomic mass of O)
= (2 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 18.02 g/mol
mass of H2O = moles of H2O * molar mass of H2O
= 1.995 mol * 18.02 g/mol
= 35.95 g
Therefore, 35.95 grams of water are formed in the reaction.
The limiting reagent is not H2O and grams H2O not equal to 28.9. If you will post your work I'll find the error.
the limiting is H2O SINCE IITS THE LEAST
COMES OUT TO BE 28.9 G H2O