First, write the balanced equation:
2KClO3−→−heat2KCl+3O22KClO3→heat2KCl+3O2
Get the MrMr of KClO3KClO3 and KClKCl(122.55 g mol−1122.55 g mol−1 and 74.6 g mol−174.6 g mol−1 respectively)
Next, find out how many moles of KClO3KClO3 are present. From calculations (n=m/Arn=m/Ar), there are 0.463 mol of KClO3KClO3 present.
From the above, we can infer that 2 mol of KClO3KClO3 decompose to give 2 mol of KClKCl. So, by the complete thermal decomposition of 0.463 mol of KClO3KClO3, we get 0.463 mol of KClKCl.
0.463 mol of KClKCl is equal to 34.5 g of KClKCl.
Hence,
34.5 g of KClKCl was produced.
