For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. kno3 → kno2 + o2

The oxidized element is, oxygen.

The reduced element is, nitrogen.

The oxidizing agent is, nitrogen.

The reducing agent is, oxygen.

Explanation:

Answer : The oxidizing element is N and reducing element is O. 

is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

 act as an oxidizing agent.

The oxidation number of N in is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in  is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number  reduced this means itself get reduced to and it can act as an oxidizing agent.

 act as a reducing agent.

The oxidation number of O in is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in is Zero (o).

Now, we conclude that the oxidation number increases this means itself get oxidized to and it can act as reducing agent.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"