Question

How do I solve part A? The reaction is 2S+3O2-->2SO3

Answer 1

We have the following balanced reaction:

`2S+3O_2\rightarrow2SO_3`

The ratio mol O2/mol S is 3/2. Now, we are given the grams of S, with this information we can calculate the moles of O2 needed in the following way:

`\begin{gathered} molO_2=\text{Given g of S }*\frac{\text{1mol S}}{\text{Molar mass of S}}*\frac{3\text{ mol O2}}{2\text{ mol S}} \\ molO_2=5.93g\text{ of S}*\frac{\text{1mol S}}{\text{32.065 g of S}}*\frac{3\text{ mol O2}}{2\text{ mol S}} \\ molO_2=0.28\text{mol} \end{gathered}`

Now, to calculate the molecules we will apply the Avogadro's number:

`\begin{gathered} \text{Molecules of O}_2=0.28\text{mol}*\frac{6.022*10^(23)molecules}{1\text{ mol}} \\ \text{Molecules of O}_2=1.67\text{ }*10^(23)molecules \end{gathered}`

So, it would be needed 1.67x10^23 molecules of O2 to react with 5.93 g of S