Question

S combines with O2, to form SO2, and SO3. If 10 g of S is mixed with 12 g of O2, What mass of SO2, and SO3, will be formed, so that neither S or O will be left at the end of reaction?​

Answer 1

Given that,

S combines with O2, to form SO2, and SO3.

The reaction will be

$S+O_{2}\Rightarrow SO_{2}$

$S+2O_{2}\Rightarrow SO_{3}+O$

Let mass of $SO_{2}$ and $SO_{3}$ are x and 22-x.

From 1st equation,

1 mol of $SO_{2}$ is formed by 1 mol of sulphate.

From 2st equation,

1 mol of $SO_{3}$ is formed by 1 mol of sulphate.

We need to calculate the value of x

Using formula of moles

$\text{mole of } SO_{2}+\text{mole of}\ SO_{3}=\text{mole of sulphur}$

$\dfrac{x}{64}+\dfrac{22-x}{80}=\dfrac{10}{32}$

$\dfrac{80x+60(22-x)}{80\times64}=\dfrac{10}{32}$

$(80x+60(22-x))\times32=80\times64\times10$

$640x+42240=51200$

$640x=51200-42240$

$640x=8960$

$x=\dfrac{8960}{640}$

$x=14$

Hence, The mass of $SO_{2}$ is 14 g.

The mass of $SO_{3}$ is 12 g.