Q-28 Potassium superoxide- KO2- is utilised in closed system breathing apparatus- Exhaled air contains CO2 nbsp-and H2O- both of which are removed and the removal of water generates oxygen for breathing by the nbsp-reaction nbsp- nbsp- 4KO2-s- - 2H2O-l- - gt- nbsp-3O2-g- - 4KOH-s- nbsp-The potassium hydroxide removes CO2from the apparatus by the reaction - nbsp-KOH -s- - CO2-g- nbsp- gt- nbsp-KHCO3-s- nbsp-a- What mass of KO2 nbsp-generates 20 gm of oxygen - nbsp-b- What mass of CO2 can be removed from the apparatus by 100 gm of KO2 nbsp- nbsp- nbsp-

Dear Student, The computation are expressed below, part #160;(a)moles #160;of #160;KO2=20 #160;g #160;O232 #160;gmol #160;O2 #215;4 #160;mo ...

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Byju's Answer

Standard XII

Chemistry

Carbon Monoxide

Q.28 Potassiu...

Question

Q.28 Potassium superoxide, $K O_{2}$ , is utilised in closed system breathing apparatus. Exhaled air contains $C O_{2}$
and $H_{2} O$ , both of which are removed and the removal of water generates oxygen for breathing by the
reaction
4 $K O_{2}$ (s) + 2 $H_{2} O$ (l) -> 3 $O_{2}$ (g) + 4KOH(s)
The potassium hydroxide removes $C O_{2}$ from the apparatus by the reaction :
KOH (s) + $C O_{2}$ (g) --> KH $C O_{3}$ (s)
(a) What mass of $K O_{2}$ generates 20 gm of oxygen ?
(b) What mass of $C O_{2}$ can be removed from the apparatus by 100 gm of $K O_{2}$ ?

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Solution

Dear Student,

The computation are expressed below,

$p a r t (a) m o l e s o f K O_{2} = \frac{20 g O_{2}}{32 \frac{g}{m o l O_{2}}} \times \frac{4 m o l K O_{2}}{3 m o l O_{2}} = 0.833 m o l g m o f K O_{2} = 0.833 m o l \times 71.097 \frac{g}{m o l} = 59.224 g m p a r t (b) m o l e s o f C O_{2} = m o l e s o f K O_{2} = \frac{100 g}{71.097 \frac{g}{m o l}} = 1.407 m o l m a s s o f C O_{2} = 1.407 m o l \times 44.01 \frac{g}{m o l} = 61.92 g m$

Regards.

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Similar questions

Potassium superoxide,

K O_{2}

, is utilised in closed system breathing apparatus. Exhaled air contains

C O_{2}

and

H_{2} O

both of which are removed and the removal of water generates oxygen for breathing by the reaction:

4 K O_{2} (s) + 2 H_{2} O (l) \to 3 O_{2} (g) + 4 K O H (s)

. The potassium hydroxide removes

C O_{2}

from the appratus by the reaction:

K O H_{s w} + C O_{2} (g) \to K H C O_{3} (s)

The mass of

K O_{2}

generating

48

g of oxygen is :

Q. A self-contained breathing' apparatus uses potassium superoxide,

K O_{2}

, to convert the carbon dioxide and water in exhaled air into oxygen, as shown by the equation:

4 K O_{2} (s) + 2 H_{2} O (g) 4 C O_{2} (g) \to 4 K H C O_{3} (s) + 3 O_{2} (g)

How many molecules of oxygen gas will be produced hem 0.0468 g of carbon dioxide that is exhaled in a typical breath?

$K O_{2}$ finds use in breathing equipment to safeguard the user to breathe in oxygen generated internally in the apparatus without being exposed to toxic fumes outside. The supply of oxygen is due to :

i.Slow decomposition of $K O_{2}$

ii.reaction of superoxide with moisture in the exhaled air

iii.reaction of $K O_{2}$ with $C O_{2}$ in the exhaled air

Q. Some older emergency oxygen masks containing potassium superoxide

(K O_{2}))

react with

C O_{2}

and water in exhaled air to produce oxygen according to the given equation.

4 K O_{2} + 2 H_{2} O + 4 C O_{2} \to 4 K H C O_{3} + 3 O_{2}

If a person exhales

0.667 g

C O_{2}

per minute, how many grams of

K O_{2}

are consumed in 5.0 minutes?

Q. Potassium superoxide (

{K O}_{2}

) is used in rebreathing gas masks to generate oxygen:

{K O}_{2} (s) + H_{2} O (l) ⟶ K O H (s) + O_{2} (g)

If a reaction vessel contains 0.158 mol

{K O}_{2}

and 0.10 mol

H_{2} O

, how many moles of

O_{2}

can be produced?

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Dear Student, The computation are expressed below, part (a)moles of KO2=20 g O232 gmol O2×4 mol KO23 mol O2=0.833 molgm of KO2=0.833 mol ×71.097 gmol = 59.224 gmpart (b)moles of CO2=moles of KO2=100 g71.097 gmol=1.407 molmass of CO2=1.407 mol×44.01 gmol=61.92 gm Regards.