Question

how many liters of oxygen gas at 21°c and 1.13 atm can be prepared by the thermal decomposition of 0.950 g kclo₃?
2 kclo₃(s) →heat 2 kcl(s) + 3 o₂(g)
2.94 l
5.67 l
10.2 l
0.248 l
1.16 l

Explanation:

Step1: Calculate moles of $KClO_3$

Molar mass of $KClO_3$: $K$ (39.10) + $Cl$ (35.45) + 3×$O$ (16.00) = 39.10 + 35.45 + 48.00 = 122.55 g/mol.
Moles of $KClO_3$ = $\frac{mass}{molar\ mass}$ = $\frac{0.950\ g}{122.55\ g/mol}$ ≈ 0.00775 mol.

Step2: Relate moles of $KClO_3$ to $O_2$ via stoichiometry

From the reaction: $2\ mol\ KClO_3$ produce $3\ mol\ O_2$.
Moles of $O_2$ = $0.00775\ mol\ KClO_3$ × $\frac{3\ mol\ O_2}{2\ mol\ KClO_3}$ ≈ 0.0116 mol.

Step3: Apply ideal gas law ($PV = nRT$) to find volume of $O_2$

Convert temperature to Kelvin: $T = 21^\circ C + 273.15 = 294.15\ K$.
$R = 0.0821\ L·atm/(mol·K)$, $P = 1.13\ atm$, $n = 0.0116\ mol$.
Rearrange $PV = nRT$ to $V = \frac{nRT}{P}$.
$V = \frac{0.0116\ mol × 0.0821\ L·atm/(mol·K) × 294.15\ K}{1.13\ atm}$ ≈ 0.248 L.

Answer:

0.248 L