Question

As with any combustion reaction, the products of combusting a hydrocarbon fuel (CxHy) with oxygen (O2) are carbon dioxide (CO2) and water (H2O).

A mass of 13.87 g for an unknown fuel was combusted in a reaction vessel containing an unknown amount of oxygen. At the end of the reaction, there still remained 13.80 g of the fuel as well as 0.1144 g of water and 0.2096 g of carbon dioxide. The oxygen was completely consumed during the reaction.

How many molecules of oxygen gas were initially present in the reaction vessel?

Express the number of molecules to three significant digits.

View Available Hint(s)

amount of O2 =    ____________   molecules

Answer 1

The full balanced reaction will be,

Here, Hydrocarbon has partially reacted. Thus, Oxygen is the limiting reagent.

Mass of CO₂ produced = 0.2096 g

Molar Mass of CO₂ = 44.01 g/mol

Moles of CO₂ , n_c= Mass / Molar mass = (0.2096 g) / (44.01 g/mol)= 0.00476255 mol

Mass of water produced = 0.1144 g

Molar Mass of water =  18.015 g/mol

Moles of water, n_w = Mass / Molar mass = (0.1144 g) / (18.015 g/mol)= 0.00635026 mol

From the balanced equation, it is clear that if moles of water is y mol and moles of CO₂ is x mol, then moles of O₂ = x+y/2

Thus, we have,

Moles of Oxygen = n_c + n_w/2 =0.00476255 mol + 0.00635026 mol/2 = 0.00793768 mol = 0.00793768 mol

1 mol contains 6.022 x 10²³ molecules.

Thus total molecules of oxygen = (0.00793768 mol) x (6.022 x 10²³ molecules/mol)= 4.78 x 10²¹ molecules (Answer)