What volume of oxygen gas is released at STP if 10.0g of potassium chlorate is decomposed? The molar mass of #KCIO_3# is 122.55 g/mol?

Question

Wyatt Atkins · Accepted Answer

#KClO_3(s) + Delta rarr KCl(s) + 3/2O_2(g)#

Approx. #3L# of dioxygen gas will be evolved.

As written, we assume that the reaction proceeds quantitatively. Moles of #KClO_3(s)# #=# #(10.0*g)/(122.55*g*mol^-1)# #=# #0.0816*mol# And thus #3/2xx0.0816*mol# dioxygen are produced, i.e. #0.122*mol#. At #"STP"#, an Ideal Gas occupies a volume of #22.4*L*mol^-1#. And thus, volume of gas produced #=# #22.4*L*mol^-1xx0.0816*mol~=3L# Note that this reaction would not work well without catalysis, typically #MnO_2#.

Avery Adams · Answer

undefined The volume of oxygen gas released is 5.60 L at STP.

Wyatt Atkins · Answer

undefined To determine the volume of oxygen gas released at STP when 10.0 grams of potassium chlorate ($KClO_3$) is decomposed, we first need to calculate the number of moles of $KClO_3$ using its molar mass.

Given:
Mass of $KClO_3$ = 10.0 g
Molar mass of $KClO_3$ = 122.55 g/mol

We use the formula:

$$
\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}
$$

Substituting the given values:

$$
\text{Number of moles} = \frac{10.0 \, \text{g}}{122.55 \, \text{g/mol}} \approx 0.0816 \, \text{moles}
$$

Now, since 2 moles of $KClO_3$ yield 3 moles of $O_2$ according to the balanced chemical equation, we can use the mole ratio to find the number of moles of $O_2$ produced.

$$
\text{Number of moles of } O_2 = \frac{3}{2} \times 0.0816 \, \text{moles} = 0.1224 \, \text{moles}
$$

To calculate the volume of $O_2$ at STP (standard temperature and pressure), we use the ideal gas law:

$$
PV = nRT
$$

Where:
$P$ = pressure (in atm)
$V$ = volume (in liters)
$n$ = number of moles
$R$ = ideal gas constant ($0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}$)
$T$ = temperature (in Kelvin)

At STP:
$P = 1 \, \text{atm}$
$T = 273.15 \, \text{K}$

Substituting the values:

$$
V = \frac{nRT}{P} = \frac{(0.1224 \, \text{moles}) \times (0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}) \times (273.15 \, \text{K})}{1 \, \text{atm}} \approx 2.67 \, \text{L}
$$

Therefore, approximately 2.67 liters of oxygen gas is released at STP when 10.0 grams of potassium chlorate is decomposed.