Explanation:
2 NO (g) + O₂ (g) ----> 2 NO₂ (g)
24.3 g of NO are reacting with 13.8 g of O₂. First we can convert the mass of theses samples into moles using their molar masses.
molar mass of O = 16.00 g/mol
molar mass of N = 14.01 g/mol
molar mass of NO = 16.00 g/mol + 14.01 g/mol
molar mass of NO = 30.01 g/mol
molar mass of O₂ = 2 * 16.00 g/mol
molar mass of O₂ = 32.00 g/mol
moles of NO = 24.3 g * 1 mol/(30.01 g)
moles of NO = 0.810 moles
moles of O₂ = 13.8 g * 1 mol/(32.00 g)
moles of O₂ = 0.431 moles
Now, to determine the limiting reactant or the excess reactant we can find the number of moles of O₂ that will react with 0.810 moles of NO and the number of moles of NO that will react with 0.431 moles of O₂.
According to the coefficients of the reaction 2 moles of NO will react with 1 mol of O₂. Let's use that relationship to find the limiting reagent.
2 moles of NO = 1 mol of O₂
moles of O₂ = 0.810 moles of NO * 1 mol of O₂/(2 moles of NO)
moles of O₂ = 0.405 moles
moles of NO = 0.431 moles of O₂ * 2 moles of NO/(1 mol of O₂)
moles of NO = 0.862 moles
We found that we need 0.405 moles of O₂ to completely react with 0.810 moles of NO. Or, we need 0.862 moles of NO to completely react with ours 0.431 moles of NO.
We can say that NO is limiting our reaction and O₂ is in excess.
Only 0.405 moles of O₂ will react with 0.810 moles of NO. But we had 0.431 moles of it. Let's find the excess.
Excess of O₂ = 0.431 moles - 0.405 moles
Excess of O₂ = 0.026 moles
Answer: 0.026 moles is the number of moles of oxygen that will be left over.
