This apparatus is used in a school laboratory to prepare oxygen from potassium chlorate. Fig. 1 shows a heated tube containing a mixture of potassium chlora...
Question 1 Report
This apparatus is used in a school laboratory to prepare oxygen from potassium chlorate. Fig. 1 shows a heated tube containing a mixture of potassium chlorate and manganese dioxide catalyst. The catalyst is not used up. Relative atomic masses: K = 39, Cl = 35.5, O = 16.
2KClO3 → 2KCl + 3O2. A mass of 12.25 g of potassium chlorate is heated completely.
(a) Give the relative formula mass of KClO3. [2]
(b) Use the mass to calculate the amount of KClO3. [1]
(c) Give the amount, in mol, of oxygen predicted by the equation. [1]
Answer Details
(a) Calculate the relative formula mass by adding the relative atomic masses of all atoms in \(\mathrm{KClO_3}\):
\[M_r(\mathrm{KClO_3})=39+35.5+(3\times16)\] [1]
\[M_r(\mathrm{KClO_3})=122.5\] [1]
(b) Use \(n=\frac{m}{M_r}\):
\[n(\mathrm{KClO_3})=\frac{12.25\text{ g}}{122.5}=0.100\text{ mol}\] [1]
(c) The equation shows that \(2\) mol of \(\mathrm{KClO_3}\) produce \(3\) mol of \(\mathrm{O_2}\). Therefore:
\[n(\mathrm{O_2})=0.100\times\frac{3}{2}=0.150\text{ mol}\] [1]
The manganese dioxide is a catalyst, so it speeds up the decomposition but is not used up and is not included in the mole calculation.
Download The App On Google Playstore
Everything you need to excel in your exams
