Consider the following reaction:
2H2S (g) + 3O2 (g) --> 2SO2 (g) + 2H2O (g)
If O2 was the excess reagent, 1.3 moles of H2S were consumed, and 18.7 g of wat were collected after the reaction has gone to completion, what is the percent yield of the equation?
What is the theoretical yield?
1.3 moles H2S x (2 moles H2O/2 moles H2S) = 1.3 moles x (1/1) = 1.3 moles H2O produced. g H2O = moles x molar mass = 1.3 moles x 18 g/mol = about 23 estimated theoretical yield. Therefore, percent yield = (18.7/theoretical yield)*100 = ?
2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
A 1.2mol
sample of H2S(g)
is combined with excess O2(g)
, and the reaction goes to completion.
Question
Which of the following predicts the theoretical yield of SO2(g)
from the reaction?
To calculate the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the maximum amount of product that could be obtained, based on stoichiometry).
Step 1: Calculate the moles of H2O produced.
From the balanced equation, we can see that 2 moles of H2S react to produce 2 moles of H2O. Therefore, if 1.3 moles of H2S were consumed, we can expect the formation of 1.3 moles of H2O.
Step 2: Convert the moles of H2O to grams.
To convert moles to grams, we need to know the molar mass of H2O. The molar mass of H2O is 18 g/mol.
1.3 moles of H2O * 18 g/mol = 23.4 g of H2O
Step 3: Calculate the percent yield.
The percent yield is calculated as (Actual Yield / Theoretical Yield) × 100%.
Actual yield = 18.7 g (given in the question)
Theoretical yield = 23.4 g (calculated in Step 2)
Percent yield = (Actual Yield / Theoretical Yield) × 100%
= (18.7 g / 23.4 g) × 100%
= 0.798 × 100%
= 79.8%
Therefore, the percent yield of the reaction is 79.8%.
To find the percent yield of a chemical reaction, you need to compare the actual yield (the amount of product obtained from the reaction) to the theoretical yield (the maximum amount of product that can be obtained based on stoichiometry).
First, let's calculate the theoretical yield of water (H2O) using the given information:
1. Determine the number of moles of water produced:
From the balanced equation, we see that 2 moles of H2S produce 2 moles of H2O. So, for every mole of H2S consumed, we should expect to produce one mole of H2O.
Since 1.3 moles of H2S were consumed, we would expect to produce 1.3 moles of H2O.
2. Calculate the molar mass of water (H2O):
H: 1.00794 g/mol x 2 = 2.01588 g/mol
O: 15.9994 g/mol
The molar mass of H2O is approximately 18.01528 g/mol.
3. Calculate the theoretical yield of water (H2O):
Theoretical yield = moles of water produced x molar mass of water
Theoretical yield = 1.3 moles x 18.01528 g/mol = 23.42 g
So, the theoretical yield of water is approximately 23.42 g.
Next, we can calculate the percent yield using the actual and theoretical yield:
Percent yield = (Actual yield / Theoretical yield) x 100%
The actual yield is given as 18.7 g.
Percent yield = (18.7 g / 23.42 g) x 100%
Percent yield = 0.7989 x 100% ≈ 79.9%
Therefore, the percent yield of the reaction is approximately 79.9%.