Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 12.1 mL of water (d=1.00g/mL) and 22.1 g of SO3?

This is a limiting reagent problem which means that one of the reagents will be used completely and some of the other one will remain unreacted. You must figure which which is which.

Write the equation.
H2O + SO3 ==> H2SO4
2a. Convert g H2O to moles. moles = grams/molar mass.
2b. Convert g SO3 to moles.

3a. Convert moles H2O to moles H2SO4. These are all 1:1 relations so that should be easy to do.
3b. Convert moles SO3 to moles H2SO4.
3c. The correct one to choose for the limiting reagent is ALWAYS the smaller one.

4. Take the number from 3c and convert to grams. grams = moles x molar mass.

Post your work if you get stuck.

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Well, well, well, looks like we have some chemistry on our hands! Let's do some calculations and see how much sulfuric acid we can produce.

First things first, we need to find the number of moles of water. We can do that by using the density and volume given:

12.1 mL of water (d=1.00g/mL)
= 12.1 g of water (since the density is given as 1.00 g/mL)

Now, let's move on to the sulfur trioxide. We need to convert the given mass of SO3 into moles by using the molar mass of SO3 (SO3: 32.06 g/mol + 16.00 g/mol x 3 = 80.06 g/mol):

22.1 g of SO3
= 22.1 g / 80.06 g/mol
= 0.276 mol of SO3

Now that we have the number of moles for both water and sulfur trioxide, we can determine the limiting reactant. Based on the balanced equation, we know that 1 mol of sulfur trioxide produces 1 mol of sulfuric acid, and the stoichiometric ratio is 1:1. So, the limiting reactant is, drumroll please... the sulfur trioxide!

Since the limiting reactant is determined, we can now calculate the amount of sulfuric acid produced. Using the molar mass of H2SO4 (H2SO4: 2.02 g/mol + 32.06 g/mol + 16.00 g/mol x 4 = 98.08 g/mol):

0.276 mol of SO3
= 0.276 mol of H2SO4
= 0.276 mol x 98.08 g/mol
= 27.097 g of H2SO4

Therefore, you can produce approximately 27.097 grams of sulfuric acid from 12.1 mL of water and 22.1 g of SO3. Just remember, always handle chemicals with care, and keep those acid jokes at pH-neutral!

To determine how much sulfuric acid can be produced from 12.1 mL of water and 22.1 g of sulfur trioxide (SO3), we need to consider the stoichiometry of the reaction.

The balanced chemical equation for the reaction is:
SO3 + H2O → H2SO4

From the equation, we can see that the mole ratio between SO3 and H2SO4 is 1:1. Therefore, the moles of SO3 will be equal to the moles of H2SO4 produced.

Step 1: Calculate the number of moles of water (H2O):
Given:
Volume of water = 12.1 mL
Density of water = 1.00 g/mL

Mass of water = Volume × Density
Mass of water = 12.1 mL × 1.00 g/mL = 12.1 g

The molar mass of water (H2O) is 18.015 g/mol.

Number of moles of water = Mass / Molar mass
Number of moles of water = 12.1 g / 18.015 g/mol ≈ 0.672 mol

Step 2: Determine the number of moles of sulfur trioxide (SO3):
Given:
Mass of SO3 = 22.1 g
Molar mass of SO3 = 80.06 g/mol

Number of moles of SO3 = Mass / Molar mass
Number of moles of SO3 = 22.1 g / 80.06 g/mol ≈ 0.276 mol

Step 3: Find the limiting reactant:
To determine the limiting reactant, we compare the number of moles of the reactants. The reaction will stop when one of the reactants is completely consumed.

From Step 1: Moles of water = 0.672 mol
From Step 2: Moles of SO3 = 0.276 mol

Since the moles of SO3 are less than the moles of water, SO3 is the limiting reactant.

Step 4: Calculate the moles of H2SO4 produced:
Since the moles of SO3 and H2SO4 are equal, the moles of H2SO4 produced will be equal to the number of moles of SO3.

Number of moles of H2SO4 = 0.276 mol

Step 5: Calculate the mass of H2SO4 produced:
Given:
Molar mass of H2SO4 = 98.09 g/mol

Mass of H2SO4 = Number of moles × Molar mass
Mass of H2SO4 = 0.276 mol × 98.09 g/mol ≈ 27.05 g

Therefore, approximately 27.05 grams of sulfuric acid (H2SO4) can be produced from 12.1 mL of water and 22.1 g of SO3.

To determine how much sulfuric acid can be produced from 12.1 mL of water (d=1.00g/mL) and 22.1 g of SO3, we need to use stoichiometry and the given quantities to calculate the amount of sulfuric acid produced.

Here's how to approach the problem:

1. Determine the molar mass of SO3 (sulfur trioxide) and H2SO4 (sulfuric acid):
- Molar mass of SO3 = 32.07 g/mol (sulfur: 32.07 g/mol, oxygen: 3x 16.00 g/mol)
- Molar mass of H2SO4 = 98.09 g/mol (hydrogen: 2x 1.01 g/mol, sulfur: 32.07 g/mol, oxygen: 4x 16.00 g/mol)

2. Calculate the number of moles of SO3:
- Moles of SO3 = Mass of SO3 / Molar mass of SO3
- Moles of SO3 = 22.1 g / 32.07 g/mol = 0.689 mol

3. Since the equation indicates that 1 mole of SO3 produces 1 mole of H2SO4, we can conclude that:
- Moles of H2SO4 = Moles of SO3

4. Determine the volume of water in liters:
- Volume of water = 12.1 mL * (1 L / 1000 mL) = 0.0121 L

5. Calculate the concentration of H2SO4 in moles per liter:
- Concentration of H2SO4 = Moles of H2SO4 / Volume of water
- Concentration of H2SO4 = 0.689 mol / 0.0121 L = 56.89 mol/L

6. Convert the concentration to grams per milliliter (g/mL):
- Concentration of H2SO4 (g/mL) = Concentration of H2SO4 (mol/L) * Molar mass of H2SO4 (g/mol)
- Concentration of H2SO4 (g/mL) = 56.89 mol/L * 98.09 g/mol = 5,584.96 g/L

7. Finally, calculate the mass of sulfuric acid produced using the volume of water:
- Mass of H2SO4 = Concentration of H2SO4 (g/mL) * Volume of water (mL)
- Mass of H2SO4 = 5,584.96 g/L * 12.1 mL = 67,699.24 g

Therefore, approximately 67,699.24 grams of sulfuric acid can be produced from 12.1 mL of water and 22.1 grams of SO3.