Question

When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O, how many grams of water are formed?

Answer 1

1) Write up a balanced Chemical Equation
C2H6+O2-->CO2+H2O
Write down the number of atoms that you have on each side of the equation.
On the Reactant Side: 2 Carbon, 6 Hydrogen and 2 Oxygen
On the Product Side: 1 Carbon, 2 Hydrogen and 3 Oxygen
Add a coefficient to the single carbon atom on the right of the equation to balance it with the 2 carbon atoms on the left of the equation.
C2H6+O2-->2CO2+H2O
Balance the hydrogen atoms next. You have 6 on the left side. So you'll need 6 on the right side. Add a coefficient of 3 to H2O.
C2H6+O2-->2CO2+3H2O
Balance the oxygen atoms.
Add a fraction of 7/2 to O2 but its better to multiply the whole equation by the smallest integer that will get rid of any fractions.
\[C2H6+7/2O2\rightarrow 2CO2+3H2O\]
2C2H6+7O2-->4CO2+6H2O
2) Determine the number of moles of both C2H6 and O2
Number of moles=Mass(g)/Molar Mass(g/mol)
Number of moles of C2H6=20.0/(12.0*2)+(1*6)=20.0g/30.0g/mol=2/3
Number of moles of O2=60.0/(16*2)=60/32=1.875
3) Determine the limiting reagent, reagent which is totally consumed when a chemical reaction takes place.
The above number of moles are the actual number of moles of each of the reactants. Use mole ratio to find the theoretical number of moles based on the actual ones.
C2H6:H2O O2:H2O
2:6 7:6
2/3 moles C2H6 * (6 moles H2O/2 moles C2H6) = 2 moles H2O
1.875 moles O2 * ( 6 moles H2O/ 7moles O2)=1.60 moles H2O
Limiting Reagent is Oxygen
Use Number of moles=Mass(g)/Molar Mass(g/mol)
Rearrange Mass(g)=Molar Mass(g/mol)*Number of moles
Mass(g)=18.0*1.60=28.8 g of H2O