List the number of each type of atom on the left side of the equation 2na3po4(aq)+2cocl2(aq)→2co3(po4)2(s )+6nacl(aq) enter your answers separated by commas (the order of the numbers is the same as the order of the elements on the left side of the equation)

Na(sodium atom) = 6 atoms , P(phosphorus atom) = 2 atoms, Oxygen(Oxygen atom) = 8 atoms , Co(cobalt atom) = 2 atoms, Cl(chlorine atom) = 4 atoms

Explanation:

2Na3PO4(aq) + 2CoCl2(aq) → 2Co3(PO4)2(s) + 6NaCl(aq)

The question ask us to list the number of the kind of atom found on the left side(reactant) of the chemical equation. The question also reiterated that the order of the numbers should be the same as the order of the elements on the left side of the equation.

The type of atoms found on the left side of the equation are as follows;

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 2 atoms

Cl(chlorine atom) = 4 atoms

The equation on the left hand side have sodium atom, phosphorus atom, oxygen atom, cobalt atom and chlorine atom.

Note: The given equation is not balanced but it can be balanced as follows;

2Na3PO4(aq) + 3CoCl2(aq) → Co3(PO4)2(s) + 6NaCl(aq)

The balanced equation have the following number of atom on the left side:

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 3 atoms

Cl(chlorine atom) = 6 atoms