0

Guest

Courses New

KO2 is used in rebreathing gas masks to generate oxygen: KO2+H2O-KOH+O2 .If a reaction vessel contains 0.158 mol KO2 and 0.10 mol H2O How many moles of O2 can be produced.

KO2 is used in rebreathing gas masks to generate oxygen: KO2+H2O-KOH+O2 .If a reaction vessel contains 0.158 mol KO2 and 0.10 mol H2O How many moles of O2 can be produced.

Grade:11

3 Answers

SANJAY SHEKHAWAT
44 Points
6 years ago
first of all see which is limiting reagent by balancing  equation . here limitining reagent is   2KO2 +H2O –2 KOH +3/2 O2 
 AS LIMITING REAGENT IS KO2 THEREFORE, MOLES OF O2  PRODUCED IS = 3/2  x  2  x  .158 =.474
Samdeep Sharma
15 Points
3 years ago
FIRST OF ALL, BALANCE THE EQUATION
2KO2 +H2O --- 2KOH +3/2 O2
KO2 IS THE LIMITING REAGENT .
SO MOLES OF O2 =3/4×0.158=0.1158 moles
MANTHAN RAO
46 Points
3 years ago
FIRST  BALANCE THE GIVEN  EQUATION

KO2 + H2O → KOH + O2 + H2O2 (unbalanced)

left side:
K = 1
O = 2 + 1 (do not add this up yet)
H = 2

right side:
K = 1
O = 1 + 2 + 2 (do not add this up yet)
H = 1 + 2 (do not add this up yet)

Always remember that in balancing equations, you are not allowed to change the subscript, but you are allowed to put coefficients before the chemical formulas.

Let's start to balance the easiest atom, in this case the H atom. Since there are 2 H atoms on the left and 3 H atoms on the right (one coming from KOH and the other two coming from H2O2), you need to find a factor that you can multiply in either one of the substances to produce a desirable number to balance.

Since KOH is the simplest of the three substances with H atoms, we'll start the balancing there.

KO2 + H2O → 2KOH + O2 + H2O2

left side:
K = 1
O = 2 + 1
H = 2

right side:
K = (1 x 2)
O = (1 x 2) + 2 + 2
H = (1 x 2) + 2

Notice that since the H atom is bonded to one K atom and one O atom, we also need to apply the coefficients to these elements. Now you have a total of 4 H atoms on the right, so to balance the H atoms on the left,

KO2 + 2H2O → 2KOH + O2 + H2O2

left side:
K = 1
O = 2 + (1 x 2)
H = 2 x 2 = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2
H = (1 x 2) + 2 = 4

Also, you have two K atoms on the right but only one K atom on the left. Thus,

2KO2 + 2H2O → 2KOH + O2 + H2O2

left side:
K = 1 x 2 = 2
O = (2 x 2) + (1 x 2)
H = 2 x 2 = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2
H = (1 x 2) + 2 = 4

Now the only thing left to balance are the O atoms. But then again if you get the sum of O atoms on both sides of the equation,

2KO2 + 2H2O → 2KOH + O2 + H2O2

left side:
K = 1 x 2 = 2
O = (2 x 2) + (1 x 2) = 6
H = 2 x 2 = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2 = 6
H = (1 x 2) + 2 = 4

2KO2 +H2O --- 2KOH +3/2 O2(BALANCED)

KO2 IS THE LIMITING REAGENT .
SO MOLES OF O2 =3/4×0.158=0.1158 moles

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More