All categories

3 years ago

Rewrite and Balance the Equation N₂ + H₂ -> NH3

Chemistry

2 answers:

snow_lady [41]3 years ago

5 0

Answer:

you said brainliest

Explanation:

Mars2501 [29]3 years ago

4 0

Answer:N

+ 3

-----> 2N

Explanation:

-----> N

Let us balance this equation by counting the number of atoms on both sides of the arrow.

-----> N

N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

on RHS

-----> 2N

N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

on LHS

+ 3

-----> 2N

N=2 , H=6 N=2 , H= 6

Answer link

Related questions

What is the chemical equation for photosynthesis?

How can I know the relative number of grams of each substance used or produced with chemical equations?

How can I know the relative number of moles of each substance with chemical equations?

How do chemical equations illustrate that atoms are conserved?

How can I know the formula of the reactants and products with chemical equations?

How can I balance this chemical equations? Potassium metal and chlorine gas combine to form...

How many types of chemical reactions exist?

How can a chemical equation be made more informative?

How can I balance this equation? ___ AlBr3 + ____ K2SO4 ---> ____ KBr + ____ Al2(SO4)3

How can I balance this equation? ____ Pb(OH)2 + ____ HCl ---> ____ H2O + ____ PbCl2

You might be interested in

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

(a) As per question, lead is oxidized and copper is reduced.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

(b) As per question, lead is reduced and copper is oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$