116

You visited us 116 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

Some older em...

Question

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 $0.667 g$ of $C O_{2}$ per minute, how many grams of $K O_{2}$ are consumed in 5.0 minutes?

10.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0757

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.07

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.38

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 5.38
The number of moles of $C O_{2}$ exhaled in 5 minutes.
$= \frac{5 \times 0.667}{44}$
$. . .$ 4 mole of $K O_{2}$ consumed by 4 mole $C O_{2}$
mass of $K O_{2} = \frac{5 \times 0.667}{44} \times 71 = 5.38 g$

Suggest Corrections

Similar questions

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. 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?

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 :

$K O_{2}$ 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. 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?

Join BYJU'S Learning Program

Some older emergency oxygen masks containing potassium superoxide (KO2)) react with CO2 and water in exhaled air to produce oxygen according to the given equation. 4KO2+2H2O+4CO2→4KHCO3+3O2 If a person exhales 0.667 g of CO2 per minute, how many grams of KO2 are consumed in 5.0 minutes?

The correct option is D 5.38The number of moles of CO2 exhaled in 5 minutes. =5×0.66744 . .. 4 mole of KO2 consumed by 4 mole CO2 mass of KO2=5×0.66744×71=5.38g

The correct option is D 5.38
The number of moles of $C O_{2}$ exhaled in 5 minutes.
$= \frac{5 \times 0.667}{44}$
$. . .$ 4 mole of $K O_{2}$ consumed by 4 mole $C O_{2}$
mass of $K O_{2} = \frac{5 \times 0.667}{44} \times 71 = 5.38 g$