Find the theoretical yield when 100 mL of 0.1 M NaI is reacted with 0.50 mL of Br2 (density: 3.10 g/mL) to produce NaBr and I2. What is the actual yield if the percent yield is 81%? (Atomic masses (g/mol): Na, 22.99; I, 126.9; Br, 79.90)
I solve these limiting reagent problems the long way by solving two simpler stoichiometry problems. Here is a step by step procedure for solving the simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Solve the first time using moles NaI (moles = M x L) and solve for moles I2 produced. Solve the second time using moles Br2 for moles I2 formed. You have two answers and of course one of them must be wrong. In limiting reagent problems, the smaller value is ALWAYS the correct one to use.
The equation is
2NaI + Br2 ==> 2NaBr + I2.
After finding the theoretical yield from the calculations above, then
theoretical yield x 0.81 = actual yield.
Post your work if you get stuck.
To find the theoretical yield, we need to calculate the number of moles of NaI and Br2 and determine which is the limiting reactant.
First, let's calculate the number of moles of NaI:
Molarity (M) = moles/volume (L)
Given:
Volume of NaI solution = 100 mL = 0.1 L
Molarity (NaI) = 0.1 M
Moles of NaI = Molarity × Volume
= 0.1 M × 0.1 L
= 0.01 moles
Next, let's calculate the number of moles of Br2:
Density = mass/volume
Given:
Volume of Br2 = 0.50 mL
Density of Br2 = 3.10 g/mL
Mass of Br2 = Density × Volume
= 3.10 g/mL × 0.50 mL
= 1.55 g
Now, let's convert the mass of Br2 to moles using its molar mass:
Molar mass (Br2) = 79.90 g/mol
Moles of Br2 = Mass/Molar mass
= 1.55 g/79.90 g/mol
= 0.0194 moles
Now, we need to determine the limiting reactant. We compare the number of moles of NaI and Br2 and use the stoichiometry of the balanced equation to see which one will react with the other in the limiting amount.
The balanced equation for the reaction is:
2NaI + Br2 -> 2NaBr + I2
From the balanced equation, we see that 2 moles of NaI react with 1 mole of Br2.
For NaI:
Moles available = 0.01 moles
For Br2:
Moles available = 0.0194 moles
We compare the moles of NaI to Br2 using the stoichiometry. Since the ratio of NaI to Br2 is 2:1, we need to use half the number of moles for NaI so that they can react in the balanced ratio.
Adjusted moles of NaI = 0.01 moles / 2
= 0.005 moles
Comparing the adjusted moles of NaI and the moles of Br2, we see that the adjusted moles of NaI are smaller. Therefore, NaI is the limiting reactant.
Now, let's calculate the theoretical yield of NaBr and I2 based on the stoichiometry of the balanced equation.
From the balanced equation, we see that:
2 moles of NaI produce 2 moles of NaBr.
Since we have 0.005 moles of NaI, the moles of NaBr produced would be the same.
Next, we determine the molar mass of NaBr:
Molar mass (NaBr) = Atomic mass (Na) + Atomic mass (Br)
= 22.99 g/mol + 79.90 g/mol
= 102.89 g/mol
The theoretical yield of NaBr can be calculated as follows:
Theoretical yield (NaBr) = Moles of NaBr × Molar mass (NaBr)
= 0.005 moles × 102.89 g/mol
= 0.514 g
To find the actual yield, we use the percent yield formula:
Percent yield = (Actual yield/Theoretical yield) × 100
Given that the percent yield is 81%, we rearrange the formula to solve for the actual yield:
Actual yield = (Percent yield/100) × Theoretical yield
= (81/100) × 0.514 g
≈ 0.417 g
Therefore, the actual yield is approximately 0.417 grams.