Question

Elemental s reacts with o2 to form so3 according to the reaction 2s+3o2→2so3a.) how many o2 molecules are needed to react with 3.50 g of s?b.) what is the theoretical yield of so3 produced by the quantities described in part a?

Answer 1

The number of molecules of
`O_2` are,
`9.88* 10^(22)`

The theoretical yield of
`SO_3` is, 8.72 grams.

Explanation :

The given balanced chemical reaction is:

`2S+3O_2\rightarrow 2SO_3`

First we have to calculate the moles of S.

`\text{Moles of }S=\frac{\text{Mass of }S}{\text{Molar mass of }S}`

Molar mass of S = 32 g/mole

`\text{Moles of }S=(3.50g)/(32g/mol)=0.109mole`

Now we have to calculate the moles of
`O_2`

From the balanced chemical reaction, we conclude that:

As, 2 mole of
`S` react with 3 mole of
`O_2`

So, 0.109 moles of
`S` react with
`(0.109)/(2)* 3=0.164` moles of
`O_2`

Now we have to calculate the number of molecules of
`O_2`

As, 1 mole of
`O_2` contains
`6.022* 10^(23)` number of
`O_2` molecules.

So, 0.164 mole of
`O_2` contains
`0.164* 6.022* 10^(23)=9.88* 10^(22)` number of
`O_2` molecules.

Thus, the number of molecules of
`O_2` are,
`9.88* 10^(22)`

Now we have to calculate the moles of
`SO_3`

From the reaction, we conclude that

As, 2 mole of
`S` react to give 2 mole of
`SO_3`

So, 0.109 moles of
`S` react to give 0.109 moles of
`SO_3`

Now we have to calculate the mass of
`SO_3`

`\text{ Mass of }SO_3=\text{ Moles of }SO_3* \text{ Molar mass of }SO_3`

Molar mass of
`SO_3` = 80 g/mol

`\text{ Mass of }SO_3=(0.109moles)* (80g/mole)=8.72g`

Thus, the theoretical yield of
`SO_3` is, 8.72 grams.