Sulfuric acid (H2SO4)is prepared commercially from elemental sulfur using the contactprocess. In a typical sequence of reactions,the sulfur is first burned:S + O2→SO2,then it is converted to SO3 using a catalyst:2SO2+O2→2SO3.The resulting SO3 is reacted with water to produce the desired product:SO3+ H2O→H2SO4. How much sulfuric acid could be prepared

from 53 moles of sulfur?
Answer in units of g

S + O2→SO2

2SO2+O2→2SO3
SO3+ H2O→H2SO4

1 mol S = 1 mol SO2
2 mol SO2 = 2 mol SO3
1 mol SO3 = 1 mol H2SO4
Therefore, 2 mols S = 2 mol H2SO4 or
1 mol S = 1 mol H2SO4

Thus 53 mols S will produce 53 mol H2SO4. Convert mols to g. g = mols x molar mass.

thank you

To determine how much sulfuric acid can be prepared from 53 moles of sulfur, we need to calculate the molecular weight of sulfuric acid (H2SO4) and then perform the appropriate calculations.

The molecular weight of sulfuric acid (H2SO4) can be calculated by adding up the atomic weights of its constituent elements:

H (hydrogen) = 1.00784 g/mol
S (sulfur) = 32.06 g/mol
O (oxygen) = 16.00 g/mol

Molecular Weight of H2SO4 = (2 * H) + S + (4 * O)
= (2 * 1.00784) + 32.06 + (4 * 16.00)
= 2.01568 + 32.06 + 64.00
= 98.07568 g/mol

Now, we can calculate the amount of sulfuric acid that can be prepared from 53 moles of sulfur:

Moles of H2SO4 = Moles of Sulfur
= 53 moles

Mass of H2SO4 = Moles of H2SO4 * Molecular Weight of H2SO4
= 53 moles * 98.07568 g/mol
= 5205.40424 g

Therefore, 53 moles of sulfur can produce 5205.40424 grams of sulfuric acid.