Question

balance the following equation by entering the correct numeric coefficients. if the coefficient is one, type in 1.
ag(s) + h₂s(g) + o₂(g) → ag₂s(s) + h₂o(l)

Explanation:

Step1: Write the unbalanced equation

$$Ag_2S(s)+H_2O(l)\to Ag(s)+H_2S(g)+O_2(g)$$

Step2: Balance the atoms

• Sulfur (S): There is 1 S atom on the left (in \(Ag_2S\)) and 1 S atom on the right (in \(H_2S\)). So S is balanced for now.
• Silver (Ag): There are 2 Ag atoms on the left (in \(Ag_2S\)) and 1 Ag atom on the right. So we put a 2 in front of \(Ag\) on the right:

$$Ag_2S(s)+H_2O(l)\to 2Ag(s)+H_2S(g)+O_2(g)$$

• Hydrogen (H): There are 2 H atoms on the left (in \(H_2O\)) and 2 H atoms on the right (in \(H_2S\)). So H is balanced for now.
• Oxygen (O): There is 1 O atom on the left (in \(H_2O\)) and 2 O atoms on the right (in \(O_2\)). Let's use a fraction. If we put \(\frac{1}{2}\) in front of \(O_2\):

$$Ag_2S(s)+H_2O(l)\to 2Ag(s)+H_2S(g)+\frac{1}{2}O_2(g)$$
But we usually prefer whole - number coefficients. Multiply the entire equation by 2 to get rid of the fraction:
$$2Ag_2S(s)+2H_2O(l)\to 4Ag(s)+2H_2S(g)+O_2(g)$$

Answer:

2 (for \(Ag_2S\)), 2 (for \(H_2O\)), 4 (for \(Ag\)), 2 (for \(H_2S\)), 1 (for \(O_2\))