In the reaction between Fe₂S₃(s) and HBr(aq), the sum of the coefficients for all reactants and products is 12.
If the reaction occurs between Fe₂S₃(s) and HBr(aq) and the equation is properly balanced with the smallest whole-number coefficients, what is the sum of the coefficients for all reactants and products?
Let's consider the unbalanced equation for the reaction between Fe₂S₃(s) and HBr(aq). This is a double displacement reaction.
Fe₂S₃(s) + HBr(aq) ⇒ FeBr₃(aq) + H₂S(g)
We will balance it using the trial and error method.
First, we will balance Fe atoms by multiplying FeBr₃ by 2.
Fe₂S₃(s) + HBr(aq) ⇒ 2 FeBr₃(aq) + H₂S(g)
Then, we will balance S atoms by multiplying H₂S by 3.
Fe₂S₃(s) + HBr(aq) ⇒ 2 FeBr₃(aq) + 3 H₂S(g)
Finally, we will get the balanced equation by multiplying HBr by 6.
Fe₂S₃(s) + 6 HBr(aq) ⇒ 2 FeBr₃(aq) + 3 H₂S(g)
The sum of the coefficients for all reactants and products is:
[tex]1 + 6 + 2 +3 = 12[/tex]
In the reaction between Fe₂S₃(s) and HBr(aq), the sum of the coefficients for all reactants and products is 12.
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