<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>
Approx. <span>3L</span> of dioxygen gas will be evolved.
Explanation:
We assume that the reaction as written proceeds quantitatively.
Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>
And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.
At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.
And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>
Note that this reaction would not work well without catalysis, typically <span>Mn<span>O2</span></span>.
Answer:
reacts to form carbondioxide
Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
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