Question

balance this equation​

Answer 1

Heyy !

Here is the answer

$H_{3}PO_{4} + NaOH \longrightarrow Na_{3}PO_{4} + H_{2}O$

STEP-1:

Elements involved in reaction - H, P, O, Na

In LHS, (the reactants)

number of elements

Hydrogen - 4

Oxygen - 5

Phosphor - 1

Sodium - 1


In RHS, (the products)

number of elements

Hydrogen - 2

Oxygen - 5

Phosphor - 1

Sodium - 3


Here, We find that Oxygen and Phosphor are balanced.

And there are 3 Sodium elements in the products of the reaction.

So we need to change the no. of molecules of NaOH in the reactants of the reaction.


STEP-2 :

Trying this, we have.

$H_{3}PO_{4} + 3NaOH$

In LHS, (the reactants)

number of elements

Hydrogen - 6

Oxygen - 7

Phosphor - 1

Sodium - 3


STEP-3:

Note no. of elements in the product side

In RHS, (the products)

number of elements

Hydrogen - 2

Oxygen - 5

Phosphor - 1

Sodium - 3

STEP-4.

Note how many extra elements has to be added for Reactants to balance the reaction

Hydrogen - 2 + 4

Oxygen - 5 + 2

Phosphor - 1 + 0

Sodium - 3 + 0

Step-5

Elements that requires the change in number in reactants side are H, and O

These two elements are combinely given in the second term of the Reactants as water.

So, we need to change the no. of molecules of $H_{2}O$ to balance the reaction.


STEP-6:

We need 6 Hydrogen and 7 Oxygen at the product side of reaction

Exclude the number of Oxygen in First term of products

7 Oxygen - 4 Oxygen = 3 Oxygen

and

6 Hydrogen.

Thus,

No. of molecules of water depends on the number of Oxygen elements availability

Thus we get 3 molecules of $H_{2}O$



Therefore,

$\fbox{H_{3}PO_{4} + 3NaOH \longrightarrow Na_{3}PO_{4} + 3H_{2}O}$


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