25.8 g of NH₃ are reacting with excess oxygen to produce 40.8 g of NO.
NH₃ + O₂ ------> NO + H₂O
The first step to find the theoretical yield is to convert the 25.8 g of NH₃ to moles of it. To do that we need the molar mass.
atomic mass of N = 14.01 amu
atomic mass of H = 1.01 amu
molar mass of NH₃ = 14.01 + 3 * 1.01
molar mass of NH₃ = 17.04 g/mol
Now we can find the number of moles of NH₃ that reacted.
number of moles of NH₃ = mass of NH₃/molar mass of NH₃
number of moles of NH₃ = 25.8 g/ (17.04 g/mol)
number of moles of NH₃ = 1.51 moles
NH₃ + O₂ ------> NO + H₂O
A reaction is like a recipe and the coefficients are the quantities. We can read it like this: 1 mol of NH₃ reacts with 1 mol of O₂ to give 1 mol of NO and 1 mol of H₂O. So we can say that 1 mol of NH₃ with excess O₂ gas will produce 1 mol of NO. Let's use that relationship to find the number of moles NO produced by 1.51 moles of NH₃.
number of moles of NO = 1.51 moles of NH₃ * 1 mol of NO/(1 mol of NH₃)
number of moles of NO = 1.51 moles
And finally we have to convert those moles into grams to determine the theoretical yield. We will need again the molar mass of NO.
atomic mass of N = 14.01
atomic mass of O = 16.00
molar mass of NO = 14.01 + 16.00
molar mass of NO = 30.01 g/mol
mass of NO = number of moles of NO * molar mass of NO
mass of NO = 1.51 moles * 30.01 g/mol
mass of NO = 45.3 g
Answer: the theoretical yield is 45.3 g of NO.
The percent yield can be found using this formula:
% yield = actual yield/ theoretical yield * 100
According to the problem the actual yield is 40.8 g of NO. So we can replace it in the formula and find the percent yield.
% yield = actual yield/ theoretical yield * 100
% yield = 40.8g /45.3 g *100
% yield = 90.1 %
Answer: the percent yield is 90.1 %.