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How many oxygen atoms are on the reactant side of this chemical equation?
2K3PO4(aq)+3Pb(NO3)2(aq)→6KNO3(aq)+Pb3(PO4)2(s)

Sagot :

whitneytr12

Answer:

The number of oxygen atoms on the reactant side of the equation is 26.

Step-by-step explanation:

The equation is:

2K₃PO₄(aq) + 3Pb(NO₃)₂(aq) → 6KNO₃(aq) + Pb₃(PO₄)₂(s)

We can see that the reaction is balanced, that is to say, the number of atoms of the reactants is equal to the number of atoms of the products.

                     

Now, we can count the number of oxygen atoms on the reactant side by multiplying the coefficients of K₃PO₄ and Pb(NO₃)₂ by the number of oxygen atoms as follows:        

[tex] O_{r} = 2*4 + 3*(3*2) = 26 [/tex]        

This is equal to the number of oxygen atoms of the products side:

[tex] O_{p} = 6*3 + 1*(4*2) = 26 [/tex]        

Therefore, the number of oxygen atoms on the reactant side of the equation is 26.

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