What is the theoretical yield of H[2]O (in grams) if 8.3 mol H[2]S were consumed in the above reaction? If 137.1g of water was actually collected after the reaction, what is the percent yield?
I have no idea to do this problem. I am not good in math at all and currently taking algebra 1 which I did not pass first semester so you can see that chemistry is killing me now. Thanks. Ty
I have no idea what the reaction is.
I am surprised your chem teacher has not counseled you yet on the math issue.
I have an IEP and they don't care and they aren't following the IEP guidelines either. I am drowning.
Did you read Bob Pursley's message? He is telling you that you didn't write the "above reaction". We can't read your mind and you wrote the problem part but not all of the question. Write the equation and we may be able to help.
2H[2}S(g) + 30[2] (g) arrow 2SO[2] (g) + 2H[2]O (g)
8.3 mol H2S used. Use the coefficients in the balanced equation to convert mols H2S to mols H2O.
That's 8.3 mol H2S x (2 mols H2O/2 mols H2S) = 8.3 x 2/2 = 8.3 mols H2O formed.
Convert that to grams.
grams H2O = mols H2O x molar mass H2O = estimated 149 g (but you need to do it more accurately) and that is the theoretical yield. The problem gives the actual yield of 137.1 g H2O.
% yield = (actual yield/theor yield)*100 = (137.1/149)*100 = ? Remember to confirm my estimates above and use YOUR numbers in the percent yield calculation and mass H2O produced.
No problem! I'll break down the problem for you step by step.
The question is asking about the theoretical yield and percent yield of water (H[2]O) in a chemical reaction.
First, let's determine the balanced chemical equation for the reaction. From the question, it seems like the reaction is between hydrogen sulfide (H[2]S) and water (H[2]O). The balanced equation looks like this:
H[2]S + H[2]O → H[2]SO[3]
According to the equation, 1 mole of H[2]S reacts to produce 1 mole of H[2]O.
The next step is to calculate the theoretical yield of water. The given information tells us that 8.3 moles of H[2]S were consumed. Since the reaction equation states a 1:1 ratio between H[2]S and H[2]O, we can conclude that the theoretical yield of water is also 8.3 moles.
To convert moles of a substance to grams, we need to know the molar mass of that substance. The molar mass of water (H[2]O) is approximately 18.015 g/mol. Therefore, the theoretical yield of water in grams can be calculated as follows:
Theoretical yield of water = 8.3 moles H[2]O x 18.015 g/mol
Now you can perform the calculation:
Theoretical yield of water = 149.4495 grams ≈ 149.45 g
So, the theoretical yield of water is approximately 149.45 grams.
Moving on to the second part of the question, we are given that 137.1 grams of water were actually collected. The percent yield is calculated by dividing the actual yield by the theoretical yield, and then multiplying by 100. The formula for percent yield is:
Percent yield = (actual yield / theoretical yield) * 100
In this case:
Percent yield = (137.1 g / 149.45 g) * 100
Now, perform the calculation:
Percent yield = 91.72%
So, the percent yield of water is approximately 91.72%.
I hope this explanation helps you understand how to solve the problem. If you have any further questions, feel free to ask!