The octane rating of gasoline is a relationship of the burning efficiency of the given gasoline mixture to the burning efficiency of octane (CgH18). Like most hydrocarbons, octane reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is
Cs H18 (g) + 02(g) CO2(g) + H20(g)
1.16 g H2 is allowed to react with 10.4 g N2, producing 2.93 g NH.
After the reaction, how much octane is left?
To solve for the amount of octane left after the reaction, we need to first balance the chemical equation:
C8H18 + 12.5 O2 → 8 CO2 + 9 H2O
Next, we need to calculate the molar masses of the compounds involved:
- Octane (C8H18) has a molar mass of approximately 114 g/mol
- Oxygen gas (O2) has a molar mass of approximately 32 g/mol
- Carbon dioxide (CO2) has a molar mass of approximately 44 g/mol
- Water (H2O) has a molar mass of approximately 18 g/mol
Now, let's calculate the moles of octane that reacted using the given mass:
10.4 g of N2 is roughly 0.75 mol, which means that 1.16 g of H2 is also about 0.75 mol.
Given the stoichiometry of the reaction, 1 mol of octane reacts with 12.5 mol of O2.
Therefore, 0.75 mol of H2 would theoretically react with 9.375 mol of O2.
Now, let's calculate the mass of 9.375 mol of O2:
9.375 mol x 32 g/mol = 300 g of O2
Given that the initial mixture had 10.4 g of N2, the total initial mixture mass is:
10.4 g + 1.16 g = 11.56 g
Therefore, the mass of octane left after the reaction can be calculated as:
11.56 g - (300 g/12.5 x 114 g/mol) = 6.13 g
So, there would be approximately 6.13 g of octane left after the reaction.