Question

what is the number of each type of atom on the right side of the equation 2Na3PO4(aq)+2CoCl2(aq)→2Co3(PO4)2(s)+6NaCl(aq)

Answer 1

Answer : The number of atoms $Co,P,O,Na\text{ and }Cl$ are, 6, 4, 16, 6, 6

Explanation :

The given balanced chemical reaction is:

$2Na_3PO_4(aq)+2CoCl_2(aq)\rightarrow 2Co_3(PO_4)_2(s)+6NaCl(aq)$

This reaction is balanced reaction because in this reaction all the atoms of individual elements are completely balanced.

From the balanced reaction we conclude that $Co_3(PO_4)_2$ and $NaCl$ are present on the right side of the equation.

Number of atoms of cobalt (Co) present on right-side of the equation = 6

Number of atoms of phosphorous (P) present on right side of the equation = 4

Number of atoms of oxygen (O) present on right side of the equation = 16

Number of atoms of sodium (Na) present on right side of the equation = 6

Number of atoms of chlorine (Cl) present on right side of the equation = 6

Hence, the number of atoms $Co,P,O,Na\text{ and }Cl$ are, 6, 4, 16, 6, 6

Answer 2

6 atoms of Cobalt.

4 atoms to Phosphate (PO4).

6 atoms to sodium.

6 atoms to Chlorine.

The coefficient is one factor that tells you how many atoms go to each atom/element, however, the subscript also influences this.

If the compound has parenthesis, then the subscript within the parenthesis remains untouched and does not affect the atoms.

Basically, to calculate the atoms is multiply the coefficient (number in front of the atom) by the subscript attached to the atom. If it's a compound such as Co3PO4, then make sure the coefficient affects the second atom as well.

Hope this helps!