N2 + 2 O2 ---> 2 NO2.
Suppose 31.25 g of N and 44.0 g of oxygen react.
A. Calculate the mass of Nitrogen dioxide formed.
B. Which reactant is the limiting reactant?
C. Calculate the mass of the excess reactant left after the reaction.
D. Find the % yield if 52.3 g of NO2 is produced
a & b.
mols N2 = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced equation convert mols N2 to mols NO2.
Do the same for mols O2 to mols NO2.\
It is likely that the mol values will be different which means one of them is not right. The correct value in limiting reagent (LR) problems is ALWAYS the smaller value and the reagent producing that value is the LR. The non-limiting reagent is the excess reagent (ER).
Using the smaller value convert that to grams. g = mol x molar mass. This is the theoretical yield(TY)
c.
Using the coefficients in the balanced equation, convert mols LR used (all of it of course) to mols used of ER. Then mols initially - mols used = mols left.
d.
%yield = (actual yield)/TY)*100 = ?
To answer these questions, we need to follow a set of steps:
Step 1: Determine the balanced chemical equation:
The balanced equation given is: N2 + 2 O2 -> 2 NO2
Step 2: Calculate the number of moles of each reactant:
To determine the number of moles, we divide the mass of each reactant by its molar mass. The molar mass of N2 is 28 g/mol, and the molar mass of O2 is 32 g/mol.
A. Calculate the mass of Nitrogen dioxide formed:
To find the mass of nitrogen dioxide (NO2) formed, we first need to calculate the number of moles of N2. Dividing the given mass of N2 (31.25 g) by its molar mass (28 g/mol) gives us:
Number of moles of N2 = 31.25 g / 28 g/mol = 1.116 moles
According to the balanced equation, 1 mole of N2 produces 2 moles of NO2. Therefore, the number of moles of NO2 produced will be twice the moles of N2:
Number of moles of NO2 = 2 * 1.116 moles = 2.232 moles
To determine the mass of NO2, we multiply the number of moles by the molar mass of NO2 (46 g/mol):
Mass of NO2 = 2.232 moles * 46 g/mol = 102.432 g
Therefore, the mass of nitrogen dioxide formed is 102.432 g.
B. Determine the limiting reactant:
To identify the limiting reactant, we need to compare the number of moles of each reactant present and their stoichiometric ratios. The limiting reactant is the one that is completely consumed, determining the maximum amount of product formed.
Number of moles of O2 = 44.0 g / 32 g/mol = 1.375 moles
According to the balanced equation, 1 mole of N2 reacts with 2 moles of O2. Thus, 1.116 moles of N2 require (2 * 1.116) = 2.232 moles of O2.
Since we have more moles of O2 (1.375) than the stoichiometrically required by N2 (2.232), the limiting reactant is N2.
C. Calculate the mass of the excess reactant left after the reaction:
To determine the mass of the excess reactant left, we need to calculate the moles of the limiting reactant (N2) consumed during the reaction.
Moles of N2 consumed = (moles of N2 initially present) - (moles of N2 reacted)
Moles of N2 consumed = 1.116 moles - 1.116 moles = 0 moles
Since all the moles of N2 are consumed, there is no excess reactant left.
D. Calculate the percentage yield:
The percentage yield is calculated by dividing the actual mass of the product obtained by the theoretical mass of the product (calculated based on the limiting reactant) and then multiplying by 100.
First, we need to calculate the theoretical mass of NO2 based on the limiting reactant (N2):
Theoretical moles of NO2 = (moles of N2) * (moles of NO2 produced per mole of N2)
Theoretical moles of NO2 = 1.116 moles * (2 moles of NO2 / 1 mole of N2) = 2.232 moles
Theoretical mass of NO2 = (theoretical moles of NO2) * (molar mass of NO2)
Theoretical mass of NO2 = 2.232 moles * 46 g/mol = 102.432 g
Percentage yield = (actual mass of NO2 produced / theoretical mass of NO2) * 100
Percentage yield = (52.3 g / 102.432 g) * 100 = 51.05%
Therefore, the percentage yield is approximately 51.05%.