The reaction consumed 0.07 g of the fuel to produce 0.1144 g of water and 0.2096 g of carbon dioxide. By calculating the mass of oxygen in these products and determining the number of moles, we find that approximately 4.77 x 10²¹molecules of oxygen gas were consumed.
To determine how many molecules of oxygen gas were initially present in the reaction vessel, we first need to calculate the mass of the fuel that was combusted. The initial mass of the fuel was 17.14 g, and after the combustion, 17.07 g remained, indicating that 0.07 g of the fuel was used.
During combustion, fuel reacts with oxygen to produce carbon dioxide and water. Given the masses of the products, we have 0.1144 g of water and 0.2096 g of carbon dioxide.
To find the mass of oxygen in the products, we need to know that the molar mass of water (H₂O) is 18.02 g/mol and carbon dioxide (CO₂) is 44.01 g/mol. Oxygen contributes 16.00 g/mol to the molar mass of both water and carbon dioxide, which means:
- The mass of oxygen in 0.1144 g of water is: (16.00 g/mol / 18.02 g/mol) ×0.1144 g = 0.1016 g of oxygen.
- The mass of oxygen in 0.2096 g of carbon dioxide is: (32.00 g/mol / 44.01 g/mol) × 0.2096 g = 0.1521 g of oxygen.
Therefore, the total mass of oxygen in the products is 0.1016 g + 0.1521 g = 0.2537 g.
Now we calculate the number of moles of oxygen used: 0.2537 g / 32.00 g/mol = 0.007925 moles of oxygen gas. Since one mole of oxygen gas (O₂) contains Avogadro's number of molecules (6.022 x 10²³ molecules/mol), we can determine the number of molecules:
0.007925 moles × 6.022 x 10²³ molecules/mol = 4.77 x 10²¹ molecules of oxygen.
So, the reaction vessel initially contained approximately 4.77 x 10²¹ molecules of oxygen gas.