Question #bed9f

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Question #bed9f

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Answer 1 · 2015-06-18T14:10:43

$Zn + 2 HCl \to {ZnCl}_{2} + H_{2}$ $"Zn" +2"HCl" -> "ZnCl"_2 + "H"_2$

Explanation:

The method I find useful when it comes to balancing a chemical equation is to draw a chart corresponding to the number of moles: (Split compounds into one unit of its own but, for polyatomic count them as one whole unit)

Unbalanced chemical equation:

$Zn + HCl \to {ZnCl}_{2} + H_{2}$ $"Zn " + " HCl" -> "ZnCl"_2 + "H"_2$

The reason you need to balance a chemical equation is because energy is conserved throughout the chemical process.

Reactant side:

$Zn = 1$ $"Zn"= 1$

$H = 1$ $"H" =1$

$Cl = 1$ $"Cl" = 1$

Product side:

$Zn = 1$ $"Zn" = 1$

$H = 2$ $"H" = 2$

$Cl = 2$ $"Cl" = 2$

You need two hydrogens and two chlorines on the left and right sides, so multiply $HCl$ $"HCl"$ by $2$ $2$

Reactant side:

$Zn = 1$ $"Zn"= 1$

$H = 1 \times 2 = 2 moles of H$ $"H" =1 xx 2 = "2 moles of H"$

$Cl = 1 \times 2 = 2 moles of Cl$ $"Cl" = 1 xx 2 = "2 moles of Cl"$

Product side:

$Zn = 1$ $"Zn" = 1$

$H = 2$ $"H" = 2$

$Cl = 2$ $"Cl" = 2$

The balanced chemical equation:

$Zn + 2 HCl \to {ZnCl}_{2} + H_{2}$ $"Zn" + 2"HCl" -> "ZnCl"_2 + "H"_2$

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Question #bed9f

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