How do you balance $A g_{2} O \to A g + O_{2}$ $Ag_2O -> Ag + O_2$ ?

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How do you balance $A g_{2} O \to A g + O_{2}$ $Ag_2O -> Ag + O_2$ ?

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Answer 1 · 2016-01-16T14:48:57

$2 A g_{2} O \to 4 A g + O_{2}$ $2Ag_2Orarr4Ag+O_2$

Explanation:

Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.

Consider the given equation:

$A g_{2} O \to A g + O_{2}$ $Ag_2OrarrAg+O_2$

Notice how there is only one $O$ $O$ on the reactants side, but two on the products side. This can be fixed by multiplying the reactant by $2$ $2$ .

$2 A g_{2} O \to A g + O_{2}$ $color(red)(2)Ag_2OrarrAg+O_2$

Now that the oxygen atoms are balanced, all that remains is the silver. There are now four $A g$ $Ag$ on the reactant side and only one on the product side, which can be remedied with a coefficient of four on the products side.

$2 A g_{2} O \to 4 A g + O_{2}$ $2Ag_2Orarrcolor(blue)(4)Ag+O_2$

This is completely balanced.

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How do you balance $A g_{2} O \to A g + O_{2}$ $Ag_2O -> Ag + O_2$ ?

1 Answer

Explanation:

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How do you balance $A g_{2} O \to A g + O_{2}$ $Ag_2O -> Ag + O_2$ ?