SO2(g) + H2O(l)= H2SO(l)
What mass of sulfur dioxide is needed to prepare 36.86 g of H2SO3(l)
28.77
Well, preparing H2SO3(l) sounds like a sulfurous challenge! But don't worry, I'm here to make it sulfuriously funny!
To find the mass of sulfur dioxide needed, we need to know the molar ratio between sulfur dioxide (SO2) and sulfurous acid (H2SO3).
The balanced equation is:
SO2(g) + H2O(l) → H2SO3(l)
From the equation, we can see that the molar ratio between SO2 and H2SO3 is 1:1. So, we can use the molar mass of H2SO3 to find the needed mass of SO2.
The molar mass of H2SO3 is:
H2SO3 = 2(1.00784 g/mol) + 1(32.065 g/mol) + 3(15.999 g/mol) = 82.07 g/mol
Now, let's calculate the mass of SO2 needed:
36.86 g H2SO3 * (1 mol H2SO3 / 82.07 g H2SO3) * (1 mol SO2 / 1 mol H2SO3) * (64.0638 g SO2 / 1 mol SO2) = ??? g SO2
Once you plug in the numbers and calculate, you'll get your answer! Just make sure to double-check the calculations, or else you might end up with a sulfurminus result.
To find the mass of sulfur dioxide needed to prepare 36.86 g of sulfuric acid (H2SO4), we need to use the balanced chemical equation and the molar masses of the compounds involved.
First, let's determine the molar mass of sulfuric acid (H2SO4):
- Hydrogen (H) has a molar mass of 1 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16 g/mol.
So, the molar mass of H2SO4 is:
(2 x 1) + 32.07 + (4 x 16) = 98.09 g/mol
Now, let's look at the balanced chemical equation:
SO2(g) + H2O(l) → H2SO4(l)
The molar ratio between SO2 and H2SO4 is 1:1. This means that 1 mole of SO2 reacts to give 1 mole of H2SO4.
To find the number of moles of H2SO4 in 36.86 g, we can use the formula:
moles = mass / molar mass
moles = 36.86 g / 98.09 g/mol ≈ 0.376 moles
Now that we have the number of moles of H2SO4, we know that we need an equal number of moles of SO2. Hence, we need 0.376 moles of SO2.
Finally, to find the mass of SO2, we multiply the number of moles by its molar mass:
mass = moles × molar mass
mass = 0.376 moles × (32.07 g/mol) ≈ 12.13 g
Therefore, approximately 12.13 grams of sulfur dioxide (SO2) is needed to prepare 36.86 grams of sulfuric acid (H2SO4).
gotta work with moles.
How many moles H2SO3 in 36.86g?
Since each mole of H2SO3 requires 1 mole of SO2, you need the same number of moles.
So, convert back to g of SO2