sulfuric acid can be prepared by a multistep process summarized as: 2 SO2 + O2 + 2 H2O--> 2 H2SO4 What mass of sulfuric acid could be produced daily by a process using 38 kg per day of sulfuric dioxide with a 70% conversion efficiency ("yield"), assuming that sulfuric dioxide is the limiting reagent

Answer: 41 kg/day

I don't know how to get this answer. someone help!

mols SO2 = 38000/molar mass SO2 = ?

mols H2SO4 produced = same as mols SO2 becaue 2 mol SO2 = 2 mol H2SO4.
Then grams H2SO4 = mols H2SO4 x molar mass H2SO4 = ? g, then convert to kg. That's if the process were 100% efficiency. It isn't so that answer x 0.70 = the amount produced at 70% efficiency.

Well, it seems like you're in a bit of a pickle. Don't worry, I'm here to help you out with a little bit of humor and some chemistry know-how!

To solve this problem, we need to find out how much sulfuric acid can be produced by the given amount of sulfur dioxide. The equation tells us that 2 moles of sulfur dioxide react with 1 mole of oxygen and 2 moles of water to form 2 moles of sulfuric acid.

First, we need to convert 38 kg/day of sulfur dioxide into moles. The molar mass of sulfur dioxide is 64 grams/mol, so we can calculate the moles of sulfur dioxide as follows:

(38 kg/day) / (64 g/mol) * (1000 g/kg) = 593.75 mol/day

Now that we know the number of moles of sulfur dioxide, we can calculate the amount of sulfuric acid produced. Since sulfur dioxide is the limiting reagent, the 70% conversion efficiency also applies to the formation of sulfuric acid.

The yield or conversion efficiency tells us that only 70% of the sulfur dioxide is converted to sulfuric acid. So, we multiply the moles of sulfur dioxide by the conversion efficiency:

593.75 mol/day * 0.70 = 415.625 mol/day

Finally, we can convert the moles of sulfuric acid into kilograms:

(415.625 mol/day) * (98 g/mol) * (1 kg/1000 g) = 40.66 kg/day

And there you have it! The mass of sulfuric acid that could be produced daily is approximately 40.66 kg/day. I guess we'll have to round that up to 41 kg/day, which sounds like a pretty decent amount. Keep in mind that this answer is based on the assumptions and information provided in the problem. Chemistry can be quite tricky, but with a little humor and perseverance, we'll get through it!

To find the mass of sulfuric acid that can be produced daily using 38 kg/day of sulfur dioxide with a 70% conversion efficiency, you can follow these steps:

Step 1: Identify the limiting reagent:
In this case, sulfur dioxide (SO2) is the limiting reagent because it is given in the question and the other reactants (O2 and H2O) are not limited.

Step 2: Calculate the molar mass of sulfur dioxide (SO2):
Sulfur (S) has a molar mass of 32.06 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Therefore, the molar mass of sulfur dioxide (SO2) is:
32.06 g/mol (S) + 2 * 16.00 g/mol (O) = 64.06 g/mol.

Step 3: Convert the mass of sulfur dioxide to moles:
Using the molar mass calculated in Step 2, we can convert the mass of sulfur dioxide (38 kg/day) to moles:
38 kg * 1000 g/kg / 64.06 g/mol = 593.2 mol.

Step 4: Determine the stoichiometry of the reaction:
The balanced equation tells us that 2 moles of sulfur dioxide (SO2) react to form 2 moles of sulfuric acid (H2SO4):
2 SO2 + O2 + 2 H2O --> 2 H2SO4.

Step 5: Calculate the theoretical yield of sulfuric acid:
Since 2 moles of sulfur dioxide (SO2) react to form 2 moles of sulfuric acid (H2SO4), and we have 70% conversion efficiency, the actual yield is 70% of the theoretical yield:
Theoretical Yield = Actual Yield / Conversion Efficiency = 593.2 mol / 0.70 = 847.4 mol.

Step 6: Convert moles to mass of sulfuric acid:
The molar mass of sulfuric acid (H2SO4) is:
2 * 1.01 g/mol (H) + 32.06 g/mol (S) + 4 * 16.00 g/mol (O) = 98.09 g/mol.
Therefore, the mass of sulfuric acid that can be produced daily is:
847.4 mol * 98.09 g/mol = 83,166.73 g/day = 83.17 kg/day.

Hence, the mass of sulfuric acid that could be produced daily by this process is approximately 83.17 kg/day.]]

To determine the mass of sulfuric acid produced daily, we need to first calculate the moles of sulfur dioxide used in the reaction.

Given:
Mass of sulfur dioxide used per day = 38 kg
Molar mass of sulfur dioxide (SO2) = 64.06 g/mol

To calculate the moles of sulfur dioxide, we can use the formula:

Number of moles = Mass / Molar mass

Number of moles of sulfur dioxide = (38 kg / 1000 g/kg) / (64.06 g/mol)

Now, since the balanced chemical equation tells us that 2 moles of sulfur dioxide reacts to form 2 moles of sulfuric acid, we know that the number of moles of sulfuric acid produced will be the same as the number of moles of sulfur dioxide.

So, the number of moles of sulfuric acid = Number of moles of sulfur dioxide

Next, we need to calculate the mass of sulfuric acid produced using the molar mass of sulfuric acid (H2SO4), which is 98.08 g/mol.

Mass of sulfuric acid produced per day = Number of moles of sulfuric acid * Molar mass of sulfuric acid

Now, since the yield or conversion efficiency is given as 70%, we need to multiply the mass of sulfuric acid produced by this factor:

Mass of sulfuric acid produced per day with 70% yield = Mass of sulfuric acid produced per day * Yield %

Finally, we can substitute the values into the formula and calculate the mass of sulfuric acid produced per day:

Mass of sulfuric acid produced per day = ((38 kg / 1000 g/kg) / (64.06 g/mol)) * (98.08 g/mol) * 0.7

Solving this equation, we find that the mass of sulfuric acid produced per day is approximately 41 kg.

Therefore, the answer is 41 kg/day.