Consider the following reaction: 2NaCl(aq) + H2O(l) -> Cl2(g) + H2(g) + 2NaOH(aq)
A 10.0g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl2 was obtained, what was
the percent yield of Cl2?
To calculate the percent yield of Cl2, we need to use the actual yield and the theoretical yield.
1. Calculate the molar mass of NaCl:
Na: 22.99 g/mol
Cl: 35.45 g/mol
NaCl: 22.99 + 35.45 = 58.44 g/mol
2. Calculate the moles of NaCl used:
Moles = mass / molar mass
Moles of NaCl = 10.0 g / 58.44 g/mol = 0.171 mol
3. From the balanced equation, we can see that the ratio of NaCl to Cl2 is 2:1. Therefore, the theoretical yield of Cl2 can be calculated as follows:
Moles of Cl2 = Moles of NaCl / 2 = 0.171 mol / 2 = 0.0855 mol
4. Calculate the theoretical yield of Cl2 using its molar mass:
Mass of Cl2 = Moles of Cl2 * molar mass of Cl2
Mass of Cl2 = 0.0855 mol * (2 * 35.45 g/mol) = 6.03 g
5. Calculate the percent yield of Cl2:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (3.85 g / 6.03 g) * 100 = 63.91%
Therefore, the percent yield of Cl2 is 63.91%.
To calculate the percent yield of Cl2 in this reaction, we first need to determine the theoretical yield of Cl2 and then compare it to the actual yield.
Given:
- Mass of sodium chloride (NaCl) = 10.0 g
- Mass of water (H2O) = 10.0 g
- Mass of Cl2 obtained = 3.85 g
1. Calculate the molar mass of NaCl:
The molar mass of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl).
Atomic mass of Na = 22.99 g/mol
Atomic mass of Cl = 35.45 g/mol
Molar mass of NaCl = (22.99 g/mol) + (35.45 g/mol) = 58.44 g/mol
2. Calculate the moles of NaCl:
To find the moles of NaCl, divide the given mass by the molar mass:
Moles of NaCl = Mass of NaCl / Molar mass of NaCl
Moles of NaCl = 10.0 g / 58.44 g/mol = 0.171 mol
3. Determine the limiting reactant:
To determine the limiting reactant, we need to compare the number of moles of NaCl with the stoichiometric ratio in the balanced equation. From the balanced equation, we see that 2 moles of NaCl react to produce 1 mole of Cl2.
Therefore, the theoretical maximum moles of Cl2 that can be produced would be half the moles of NaCl used.
Theoretical moles of Cl2 = 0.171 mol / 2 = 0.0855 mol
4. Calculate the theoretical yield of Cl2:
To find the theoretical yield of Cl2 in grams, multiply the moles of Cl2 by the molar mass of Cl2:
Theoretical yield of Cl2 = Theoretical moles of Cl2 * Molar mass of Cl2
Theoretical yield of Cl2 = 0.0855 mol * 70.91 g/mol = 6.07 g
5. Calculate the percent yield of Cl2:
Percent Yield = (Actual yield / Theoretical yield) * 100
Percent Yield = (3.85 g / 6.07 g) * 100 = 63.5%
Therefore, the percent yield of Cl2 in this reaction is 63.5%.
This is a limiting reagent problem; I know that because amounts for BOTH reactants are given. I do these problems by solving two stoichiometry problems. The first, use 10 g water (ignore the NaCl), convert to moles H2O, convert to moles Cl2 using the coefficients in the balanced equation. Then use 10 g NaCl(ignore the H2O) and convert to moles NaCl, use the coefficients in the balanced equation to convert moles NaCl to moles Cl2. You will obtain two answers for moles Cl2 and the correct one in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. Use the smaller value and convert to grams. g = moles x molar mass.
That is the theoretical yield.
%yield = (actual yield/theoretical yield)*100 = ?