# balancing equations and counting atoms?

how do u balance them if there was like 2 O's or anything in 1 side for example :_Mg+_HNO3=_Mg(NO3)2+_H2O

also how do i count atom if there was a number infront example :4Al2(SO4)3 or 3(NH4)2SO3

please help!

### 2 Answers

- Seamus OLv 79 years agoFavorite Answer
For the first one:

The reaction should be:

Mg + HNO3 → Mg(NO3)2 + H2

Balancing oxygens ... there are 6 on the RHS so need the same on the LHS:

Mg + 2HNO3 → Mg(NO3)2 + H2

Now Mg is balanced ... 1 on each side

H is balanced ... 2 on each side

and N is balanced ... 2 on each side

so the balanced equation is Mg + 2HNO3 → Mg(NO3)2 + H2

The number in front (the coefficient) of a formula multiplies everything in the formula by that number

subscripts (after an atom or group of atoms in parentheses) only refers to the atom or the group of atoms in parentheses

so for Al2(SO4)3

Number of Al atoms = 2

Number of SO4 ions = 3 → so number of S atoms = 3 and number of O atoms = 4 x 3 = 12

BUT they want an atom count for 4 Al2(SO4)3 ... the coefficient 4 multiplies all the atoms in Al2(SO4)3 by 4 ... so:

Number of Al atoms = 2 x 4 = 8

Number of S atoms = 3 x 4 = 12

Number of O atoms = 4 x 3 x 4 = 48

For 3(NH4)2SO3:

Number of N atoms = 2 x 3 = 6

Number of H atoms = 4 x 2 x 3 = 24

Number of S atoms = 1 x 3 = 3

Number of O atoms = 3 x 3 = 9

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- luvmath03Lv 59 years ago
First part left side must be multiple of 3 for oxygen.....current 7 on right....

I am just not seeing it.....are you sure about the products?

second part is easy.....

total 8 Al or 12 (SO4) 12 S 48 O

6 NH4 6 N 24 H

3 SO3 3 S 9 O

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