Based on the following chemical equation answer the question below show all your calculation
The balanced chemical reaction of iodine with hydrogen sulfide h2s(g)+i2(s)-----2HI(g) + S(s)
(a) Based on the above chemically balanced equation, how many grams of HI should be produced from the reaction of 25.0g of H2S when there is excess amount of I2?
(b) Actually 30.0g of HI is produced in this reaction what is the percent yield of HI?
I worked this one or one almost the same yesterday. Look that up and follow the steps. The same applies to your posts just below.
It's not clear to me at all. Can you explain it again please
In the future, try to avoid using general statements that a problem is not clear to you. Be specific. What did you not understand? What is troubling about the process? As it stands all I can do is rework the problem and if you didn't understand it the first time it the chances are poor for the secolnd. It just takes up your time and my time to repeat a process that didn't work. However, I'll go through it in a little more detail.
These stoichiometry problems are a four step process.
1. Write and balance the equation.
H2S + I2 ==> 2HI + S
2. Convert 25.0 g H2S to mols. mols = grams/molar mass.
mols H2S = 25.0/about 34 = about0.735 but you need to confirm that with a more accurate number.
3. Using the coefficients in the balanced equation, convert mols H2S to mols HI.
0.735 mols H2S x (2 mols HI/1 mol H2S) = 0.735 x 2/1 = 1.47
4. Now convert mols HI to grams. g = mols x molar mass.
g HI = 1.47 x 128 = 188 grams HI. This is the theoretical yield and is where most of these problems stop.
5. This one tells you the actual yield is 30.0 g in the b part and want to know the percent yield.
%yield = (actual yield/theo yield)*100
= (30/188)* 100 = about 16% yield.
To calculate the grams of HI produced in the reaction and the percent yield, we need to follow these steps:
Step 1: Convert the given mass of H2S to moles.
Step 2: Use the balanced equation to determine the mole ratio between H2S and HI.
Step 3: Convert the moles of HI to grams.
Step 4: Calculate the percent yield using the experimental and theoretical values.
Let's calculate each part of the question:
(a) How many grams of HI should be produced from the reaction of 25.0g of H2S when there is an excess amount of I2?
Step 1: Convert the given mass of H2S to moles.
Molar mass of H2S = 1.0079 g/mol + 32.065 g/mol = 34.073 g/mol
Given mass of H2S = 25.0 g
Moles of H2S = 25.0 g / 34.073 g/mol ≈ 0.7349 mol
Step 2: Use the balanced equation to determine the mole ratio between H2S and HI.
From the equation, we can see that 1 mole of H2S reacts to form 2 moles of HI.
Moles of HI = 2 * 0.7349 mol = 1.4698 mol
Step 3: Convert the moles of HI to grams.
Molar mass of HI = 1.0079 g/mol + 126.9045 g/mol = 127.9124 g/mol
Grams of HI = 1.4698 mol * 127.9124 g/mol ≈ 187.763 g
Therefore, approximately 187.763 grams of HI should be produced.
(b) Calculate the percent yield of HI when 30.0g of HI is produced in this reaction.
Given mass of HI (experimental value) = 30.0 g
Percent yield is calculated by dividing the actual yield (experimental value) by the theoretical yield (calculated value) and multiplying by 100.
Theoretical yield of HI = 187.763 g (calculated in part a)
Percent yield = (30.0 g / 187.763 g) * 100 ≈ 15.99%
Therefore, the percent yield of HI in this reaction is approximately 15.99%.