Question

How many grams of oxygen can be produced from the decomposition of 100. g of kclo3?

Answer 1

100 g of KClO3 @ 122.55 g/mol = 0.816 moles of KClO3
   by the reaction
 2 KClO3 --> 2 KCl & 3 O2

   0.816 moles of KClO3 @ 3 moles O2 / 2 moles KClO3 = 1.224 moles of O2 can be made using molar mass
 1.224 moles of O2 @ 32.0 g/mol =
 39.2 grams of O2 can be made

Answer 2

Answer:

39.168g of $O_{2}$

Explanation:

Step 1: Determine the number of moles of $KClO_{3}$

$n=\frac{m}{M}$

where 'n' is the number of moles, m is the mass of compound and M is the molar mass of compound

$n_{KClO_{3}}=\frac{m_{KClO_{3}}}{M_{KClO_{3}}}$

$n_{KClO_{3}}=\frac{100.0}{39+35.5+3(16)}$

$n_{KClO_{3}}=\frac{100.0}{122.5}$

$n_{KClO_{3}}=0.816 mol$

Step 2: Determine number of moles of $O_{2}$ produced per mole of $KClO_{3}$ decomposed

$2KClO_{3} -> 3O_{2} + 2KCl$

1 mole of $KClO_{3}$ will produce 1.5 moles of $O_{2}$

$n_{O_{2}} = 1.5 * n_{KClO_{3}}$

$n_{O_{2}} = 1.5 * 0.816$

$n_{O_{2}} = 1.224 mol$

Step 3: Determine mass of $O_{2}$

$m_{O_{2}}=n_{O_{2}}M_{O_{2}}$

$m_{O_{2}}=1.224*2(16)$

$m_{O_{2}}=39.168 g$