What’s the answer to this question because I’ve look quite literally everywhere for it

To find the mass of ethane [tex]\bf{C_{2}H_{4} }[/tex] that burns in the reaction, we need to calculate the mass of ethane consumed based on the mass of the products formed.

From the balanced chemical equation:

1 mole of [tex]\bf{C_{2}H_{4} }[/tex] reacts with 3 moles of O2 to produce 2 moles of CO2 and 2 moles of H2O.

So the molar masses are:

[tex]\bf{C_{2}H_{5} }[/tex] : 2×12.01 + 4×1.01 = 28.05 g/mol
[tex]\bf{O_{2} }[/tex]: 2×16.00 = 32.00 g/mol
[tex]\bf{CO_{2} }[/tex]: 12.01 + 2×16.00 = 44.01 g/mol
[tex]\bf{H_{2}O }[/tex]: 2×1.01 + 16.00 = 18.02 g/mol

Using the given masses:

Initial mass of ethane = ?g
Initial mass of oxygen = 17.11g

Final masses of products:

CO2 = 15.69g
H2O = 6.42g

Now, let's calculate the moles of oxygen consumed, which is the limiting reactant:

Moles of [tex]\bf{O_{2} }[/tex] = Mass of [tex]\sf{O_{2} }[/tex] / Molar mass of [tex]\sf{O_{2} }[/tex] [tex]\sf{\frac{17.11g}{32.00 g/mol}~ }[/tex] ≈ 0.535 mol

According to the balanced equation, 1 mole of ethane reacts with 3 moles of oxygen.

So, the moles of ethane consumed would be:

Moles of [tex]\bf{C_{2}H_{4} }[/tex] consumed = [tex]\sf{\frac{0.535~mol~O_{2}}{3~ mol ~O_{2}/mol }~ }[/tex] ≈ 0.178 mol

Now, let's convert moles of ethane to grams:

Mass of [tex]\bf{C_{2}H_{4} }[/tex] consumed = Moles of [tex]\sf{C_{2}H_{4} }[/tex] consumed * Molar mass of C2H4

Mass of [tex]\bf{C_{2}H_{4} }[/tex] consumed ≈ (0.178 mol) × (28.05 g/mol) ≈ 5.00g

Therefore, approximately 5.00g of ethane burns in this reaction.