To determine how many O2 molecules are needed to react with 3.16 g of S, we first need to convert the mass of S into moles.
The molar mass of S is 32.06 g/mol.
Number of moles of S = mass of S / molar mass of S
Number of moles of S = 3.16 g / 32.06 g/mol = 0.0986 mol
According to the balanced equation, the ratio of moles of S to moles of O2 is 2:3. Therefore, the number of moles of O2 needed can be calculated as:
Number of moles of O2 = (2/3) * Number of moles of S
Number of moles of O2 = (2/3) * 0.0986 mol = 0.0657 mol
Now, we can convert the number of moles of O2 into molecules.
Avogadro's number tells us that 1 mole of any substance contains 6.022 x 10^23 molecules.
Number of O2 molecules = Number of moles of O2 * Avogadro's number
Number of O2 molecules = 0.0657 mol * 6.022 x 10^23 molecules/mol = 3.95 x 10^22 molecules
Therefore, 3.95 x 10^22 O2 molecules are needed to react with 3.16 g of S.
To find the theoretical yield of SO3 produced by 3.16 g of S, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of S react to form 2 moles of SO3.
Number of moles of SO3 produced = (2/2) * Number of moles of S
Number of moles of SO3 produced = (2/2) * 0.0986 mol = 0.0986 mol
The molar mass of SO3 is 80.06 g/mol.
Mass of SO3 produced = Number of moles of SO3 produced * molar mass of SO3
Mass of SO3 produced = 0.0986 mol * 80.06 g/mol = 7.889 g
Therefore, the theoretical yield of SO3 produced by 3.16 g of S is 7.889 g.
Elemental SS reacts with O2O2 to form SO3SO3 according to the
reaction
2S+3O2→2SO3
How many O2 molecules are needed to react with 3.16 g of S?
What is the theoretical yield of SO3 produced by 3.16 g
1 answer