Question

Given the following equation: 2KClO3 → 2KCl + 3O2How many moles of O2 can be produced by letting 480.5 grams of KClO3 react?

Answer 1

Let's start by calculating the molar mass of KClO₃:

`\begin{gathered} M_(KClO_3)=1\cdot M_K+1\cdot M_(Cl)+3\cdot M_O \\ M_(KClO_3)=(1\cdot39.0983+1\cdot35.453+3\cdot15.9994)g/mol \\ M_(KClO_3)=(39.0983+35.453+47.9982)g/mol \\ M_(KClO_3)=122.5495g/mol \end{gathered}`

Now, we can convert the mass of KClO₃ to number of moles of KClO₃:

`\begin{gathered} M_{KClO_(3)}=\frac{m_(KClO_3)}{n_{KClO_(3)}}_{} \\ n_(KClO_3)=\frac{m_(KClO_3)}{M_{KClO_(3)}}=(480.5g)/(122.5495g/mol)=3.9208\ldots mol \end{gathered}`

Since 2 moles of KClO₃ produce 3 moles of O₂, we have:

`\begin{gathered} (n_(KClO_3))/(2)=(n_(O_2))/(3)_{} \\ n_(O_2)=(3)/(2)n_(KClO_3)=(3)/(2)\cdot0.9208\ldots mol=5.8812\ldots mol\approx5.881mol \end{gathered}`

So, it will produce approximately 5.881 mol of O₂,