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Fiesta28 [93]
3 years ago
9

What strategies did you use when you played the balancing chemical equations game? Which atoms were the easiest to start examini

ng to try to balance the equations? And did it require trial and error?
Chemistry
2 answers:
alexandr1967 [171]3 years ago
7 0

<u>Answer:</u>

Free element is balanced at last.

We start to balance which occurs in fewer places,  first.

Balancing requires Trial and error method

<u>Balancin</u>g is making the number of atoms of each element same on both the sides  (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript  

For example 5 Ca_1 Cl_2 contains

5 × 1 = 5,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example 3Ca_3 (P_1 O_4 )_2 we multiply coefficient \times subscript \times number outside the bracket.......... to find the number of atoms  

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9, Ca atoms  

3 × 1 × 2 = 6, P atoms  

3 × 4 × 2 = 24, O atoms are present.

So

Let us balance the equation given

BaCl_2  + H_2 SO_4  > BaSO_4  + HCl

(Unbalanced)

Reactant side - Number of atoms of each element - Product side

1  - Ba - 1

2 - Cl - 1

2 - H - 1

1  - S - 1

4 - O - 4

So we see here H and Cl are not balanced

So by changing the coefficient of HCl as 2 we will have  

BaCl_2+ H_2 SO_4  > BaSO_4+2HCl

(Balanced)  

Reactant side - Number of atoms of each element - Product side

1  - Ba - 1

2 - Cl - 2

2 - H - 2

1  - S - 1

4 - O - 4

So we see the number of atoms of each element is same on both the sides.

Hence the equation is balanced !!!!!!!!!!

Yakvenalex [24]3 years ago
3 0

<u>Answer:</u>

<em>Latest take an example to understand how </em><em>balancing of chemical reaction</em><em> is done that is assuming the reaction between iron as well as </em><em>oxygen which reacts to form rust.</em>

<u>Explanation:</u>

For this we would simply right the reactant and product that is expected. Then we would see the number of molecules of each element present on either side.

If in the reaction the element’s molecules are the same on both sides then the reaction would be correct and if not then we have to apply trial and error method to balance the equation such that the number of molecules of each element is equal on both sides of the reaction.

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