When solutions of H2SO4 and NaOH react, the balanced molecular equation is: H2SO4 (aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O (I)
How much Na2SO4 is obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH?
H2SO4 (aq) + 2 NaOH(aq) ==> Na2SO4(aq) + 2 H2O (I)
This is a limiting reagent (LR) problem. You know that when quantities are given for BOTH reactants. I'm estimating the numbers so you should redo all of them.
mols H2SO4 = g/molar mass = 4/98 = about 0.04
mols NaOH = 4/40 = 0.10
If H2SO4 and excess NaOH were used then Na2SO4 produced would be 0.04 x (1 mol Na2SO4/1 mol H2SO4) = 0.04 mols Na2SO4.
If NaOH and excess H2SO4 were used then Na2SO4 produced would be
0.10 x (1 mol Na2SO4/2 mols NaOH) = 0.05 mols Na2SO4
In LR problems the smaller number always is the correct value so 0.04 mols Na2SO4 will be produced. Convert to grams = mols Na2SO4 x molar mass Na2SO4. Remember to recalculate all of the numbers.
BTW, I don't see much organic chemistry in this problem.
Oh, we've got a funny situation here! It's chemistry comedy time! So, let's find out how much Na2SO4 is produced when 4.00 grams of H2SO4 and 4.00 grams of NaOH try to have a reaction.
First, we need to figure out the number of moles of each compound. For H2SO4, we can use its molar mass (98.09 g/mol) to calculate that we have 4.00 g / 98.09 g/mol = 0.0408 mol of H2SO4.
Now, let's do the same for NaOH. Its molar mass is 40.00 g/mol, so we have 4.00 g / 40.00 g/mol = 0.100 mol of NaOH.
Since the balanced equation tells us that the ratio of H2SO4 to Na2SO4 is 1:1, we can conclude that 0.0408 mol of H2SO4 will also produce 0.0408 mol of Na2SO4.
Finally, to find the mass of Na2SO4, we multiply the number of moles (0.0408 mol) by its molar mass (142.04 g/mol) to get 5.81 grams of Na2SO4.
So, after all the calculations, we end up with 5.81 grams of Na2SO4. Ta-da! Chemistry jokes, always full of surprises.
To find the amount of Na2SO4 obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH, we need to follow these steps:
Step 1: Convert the given masses of H2SO4 and NaOH to moles.
- The molar mass of H2SO4 is 98.09 g/mol.
- The molar mass of NaOH is 40.00 g/mol.
- Using the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass, we can calculate:
- Moles of H2SO4 = 4.00 g / 98.09 g/mol
- Moles of NaOH = 4.00 g / 40.00 g/mol
Step 2: Check the stoichiometric ratio between H2SO4 and Na2SO4.
- From the balanced equation, we see that it is 1:1, meaning one mole of H2SO4 reacts with one mole of Na2SO4.
Step 3: Determine the limiting reactant.
- To do this, we compare the number of moles of H2SO4 to NaOH.
- As the stoichiometric ratio is 1:1, whichever reactant has a lower number of moles is the limiting reactant.
- Let's compare the number of moles:
- Moles of H2SO4 = 4.00 g / 98.09 g/mol = 0.0407 mol
- Moles of NaOH = 4.00 g / 40.00 g/mol = 0.100 mol
Since we have less moles of H2SO4 (0.0407 mol) than NaOH (0.100 mol), H2SO4 is the limiting reactant.
Step 4: Calculate the moles of Na2SO4 reacted.
- As we know from the stoichiometric ratio, 1 mole of H2SO4 reacts with 1 mole of Na2SO4.
- Since H2SO4 is the limiting reactant, we can conclude that 0.0407 mol of Na2SO4 reacts.
Step 5: Convert the moles of Na2SO4 to grams.
- The molar mass of Na2SO4 is 142.04 g/mol.
- Using the formula m = n x M, where m is the mass, n is the number of moles, and M is the molar mass, we can calculate:
- Mass of Na2SO4 = 0.0407 mol x 142.04 g/mol
The mass of Na2SO4 obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH is approximately 5.78 g.
To determine the amount of Na2SO4 obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH, we need to use mole-to-mole ratios and stoichiometry.
Step 1: Calculate the number of moles of H2SO4 and NaOH.
To find the moles, divide the given mass by the molar mass of each compound.
Molar mass of H2SO4:
2 * atomic mass of H (2 * 1.01 g/mol) + atomic mass of S (32.07 g/mol) + 4 * atomic mass of O (4 * 16.00 g/mol) = 98.09 g/mol
Number of moles of H2SO4 = mass / molar mass = 4.00 g / 98.09 g/mol ≈ 0.0408 mol
Molar mass of NaOH:
atomic mass of Na (22.99 g/mol) + atomic mass of O (16.00 g/mol) + atomic mass of H (1.01 g/mol) = 39.99 g/mol
Number of moles of NaOH = mass / molar mass = 4.00 g / 39.99 g/mol ≈ 0.1000 mol
Step 2: Determine the limiting reagent.
The limiting reagent is the one that is completely consumed in the reaction. To find the limiting reagent, compare the mole ratios from the balanced equation.
From the balanced equation, the mole ratio of H2SO4 to NaOH is 1:2. This means that 1 mol of H2SO4 reacts with 2 mol of NaOH.
Since the mole ratio of H2SO4 to NaOH is 1:2, we need twice as many moles of NaOH to react completely with the moles of H2SO4.
Since we have 0.0408 mol of H2SO4 and 0.1000 mol of NaOH, we can calculate the expected amount of moles for NaOH if it were the limiting reagent: 2 * 0.0408 mol ≈ 0.0816 mol
Comparing the actual amount of NaOH (0.1000 mol) to the expected limiting amount (0.0816 mol), we can see that NaOH is in excess. Therefore, H2SO4 is the limiting reagent.
Step 3: Calculate the moles of Na2SO4 formed.
Again, using the mole ratio from the balanced equation, we can determine the moles of Na2SO4 formed.
From the balanced equation, the mole ratio of H2SO4 to Na2SO4 is 1:1. This means that 1 mol of H2SO4 reacts to produce 1 mol of Na2SO4.
Since we have 0.0408 mol of H2SO4, we expect to form 0.0408 mol of Na2SO4.
Step 4: Convert moles to grams.
To calculate the mass of Na2SO4 produced, multiply the moles of Na2SO4 by its molar mass.
Molar mass of Na2SO4:
2 * atomic mass of Na (2 * 22.99 g/mol) + atomic mass of S (32.07 g/mol) + 4 * atomic mass of O (4 * 16.00 g/mol) = 142.04 g/mol
Mass of Na2SO4 = moles of Na2SO4 * molar mass = 0.0408 mol * 142.04 g/mol ≈ 5.83 g
Therefore, approximately 5.83 g of Na2SO4 is obtained when 4.00 g of H2SO4 reacts with 4.00 g of NaOH.