Question

3. How much KCIO3 must be decomposed to provide 400. mL of oxygen gas collected at STP? KCI (s)+ 02 (g) KCIO3 (s)

Answer 1

The balanced decomposition reaction of KClO₃ is as follows:

2KClO₃(s) $\rightarrow$ 2KCl(s) + 3O₂(g)

At Standard temperature and pressure(STP),

1 mol = 22.4 L of an ideal gas.

Use the conversion factor 1 mol = 22.4 L and the given volume of oxygen gas to determine the moles as follows:

= 400 mL x (1 L /1000 mL) x (1 mol / 22.4 L)

= 0.01786 mol O₂

Use the moles of O₂ and the mole ratio from the balanced chemical reaction and determine the number of moles of KClO₃ as follows:

= 0.01786 mol O₂ x (2 mol KClO₃ / 3 mol O₂)

= 0.0119 mol KClO₃

Convert the moles of KClO₃ to grams as follows:

The molar mass of KClO₃ = 122.55 g/mol

=0.0119 mol KClO₃ x ( 122.55 g KClO₃ /1 mol KClO₃)

=1.46 g KClO₃

Thus, 1.46 g KClO₃ must be decomposed to provide 400. mL oxygen gas collected at STP.