How many grams of CO2 could be produced from the combustion of 0.0713 mol. of C2H4 in the presence of 0.156 mol. of O2?

C2H4 + 3O2 --> 2CO2 + 2H2O ideally

so for .0713 mol C2H4 we need 3*.0713 = .213 mol O2
so we do not have enough O2
so .156 mol O2 is limiting
for .156 mol O2 we get
(2/3).156 mol of CO2
or
.104 mol of CO2
C = 12 g/mol
O2 = 32 g/mol
so
CO2 = 44 g/mol

44 * .104 = 4.58 grams of CO2

To find the number of grams of CO2 produced from the combustion of C2H4 (ethylene), we need to use the balanced chemical equation for the reaction:

C2H4 + O2 → CO2 + H2O

The balanced equation tells us that one mole of C2H4 reacts with three moles of O2 to produce two moles of CO2.

Given that we have 0.0713 mol of C2H4 and 0.156 mol of O2, we can find the limiting reactant. The limiting reactant is the reactant that will completely react, limiting the amount of product formed.

To determine the limiting reactant, we compare the stoichiometric ratios between the reactants and the product. In this case, we need to compare the moles of C2H4 and O2 to determine which one reacts in a 1:3 ratio (since three moles of O2 are needed for every mole of C2H4):

For C2H4:
0.0713 mol C2H4 × (3 mol O2 / 1 mol C2H4) = 0.2139 mol O2 required

For O2:
0.156 mol O2

Since the actual amount of O2 (0.156 mol) is greater than the amount required (0.2139 mol), O2 is in excess, and C2H4 is the limiting reactant.

Now that we know C2H4 is the limiting reactant, we can calculate the moles of CO2 produced using the stoichiometry of the balanced equation:

0.0713 mol C2H4 × (2 mol CO2 / 1 mol C2H4) = 0.1426 mol CO2

Finally, to convert moles of CO2 to grams, we use the molar mass of CO2 which is approximately 44.01 g/mol:

0.1426 mol CO2 × 44.01 g/mol = 6.279 g CO2

Therefore, the combustion of 0.0713 mol of C2H4 would produce approximately 6.279 grams of CO2.