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Question

# Complete and balance the following equation:$\mathrm{CuO}+\mathrm{HCl}\left(dil\mathit{.}\right)\to$

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Solution

## When metallic oxide reacts with Hydrochloric acid ($\mathbf{HCl}$) it forms its respective Chloride along with water.General reaction between Metal oxide and Hydrochloric acid is given below : $\mathrm{Metal}+\mathrm{Hydrochloric}\to \mathrm{Metal}+\mathrm{Water}\phantom{\rule{0ex}{0ex}}\mathrm{oxide}\mathrm{acid}\left(dil\mathit{.}\right)\mathrm{chloride}$Step 1: Reacting the given molecules:Copper(II) oxide ($\mathbf{CuO}$) react with Hydrogen chloride ($\mathbf{HCl}$) it produces Copper(II) chloride (${\mathbf{CuCl}}_{\mathbf{2}}$) and water.Complete reaction between Copper (II) Oxide $\left(\mathrm{CuO}\right)$ and Hydrochloric acid $\left(\mathbf{HCl}\right)$ is $\underset{\left(\mathrm{Copper}\left(\mathrm{II}\right)\mathrm{oxide}\right)}{\mathrm{CuO}\left(s\right)}+\underset{\left(\mathrm{Hydrochloric}\mathrm{acid}\right)}{\mathrm{HCl}\left(aq\right)}\to \underset{\left(\mathrm{Copper}\left(\mathrm{II}\right)\mathrm{Chloride}\right)}{{\mathrm{CuCl}}_{2}\left(aq\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_{2}\mathrm{O}\left(aq\right)}\phantom{\rule{0ex}{0ex}}$Step 2: Balancing the chemical equationThe obtained chemical reaction is unbalanced, to balance it we first count the number of molecules on both sides of the reaction.On the reactant side: 1 Copper and 1 Oxygen are present in the Copper oxide.Similarly, 1 Hydrogen and 1 Chlorine are present in Hydrogen chloride.On the product side: 1 Copper and 2 Chlorine are present in Copper chloride Similarly, 2 Hydrogen and 1 oxygen are present in the water molecule.Step 3: Balancing the reactant molecules by multiplying 2 by Hydrogen chloride.The chemical reaction is as follows: $\underset{\left(\mathrm{Copper}\mathrm{oxide}\right)}{\mathrm{CuO}\left(s\right)}+\underset{\left(\mathrm{Hydrogen}\mathrm{chloride}\right)}{2\mathrm{HCl}\left(aq\right)}\to \underset{\left(\mathrm{Copper}\mathrm{chloride}\right)}{{\mathrm{CuCl}}_{2}\left(aq\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_{2}\mathrm{O}\left(aq\right)}\phantom{\rule{0ex}{0ex}}$Therefore, after multiplication we have an equal number of molecules on both sides of the reaction.Step 4: Balanced chemical equation is as follows:The balanced chemical equation: $\underset{\left(\mathrm{Copper}\mathrm{oxide}\right)}{\mathrm{CuO}\left(s\right)}+\underset{\left(\mathrm{Hydrogen}\mathrm{chloride}\right)}{2\mathrm{HCl}\left(aq\right)}\to \underset{\left(\mathrm{Copper}\mathrm{chloride}\right)}{{\mathrm{CuCl}}_{2}\left(aq\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_{2}\mathrm{O}\left(aq\right)}\phantom{\rule{0ex}{0ex}}$The total number of molecules on both sides of the reaction is equal i.e. on the reactant and product side.On the reactant side: 1 Copper and 1 Oxygen are present in the Copper oxide.Similarly, 2 Hydrogen and 2 Chlorine are present in Hydrogen chloride.On the product side: 1 Copper and 2 Chlorine are present in Copper chloride Similarly, 2 Hydrogen and 1 oxygen are present in the water molecule.Hence the balanced chemical equation is $\underset{\left(\mathrm{Copper}\mathrm{oxide}\right)}{\mathrm{CuO}\left(s\right)}+\underset{\left(\mathrm{Hydrogen}\mathrm{chloride}\right)}{2\mathrm{HCl}\left(aq\right)}\to \underset{\left(\mathrm{Copper}\mathrm{chloride}\right)}{{\mathrm{CuCl}}_{2}\left(aq\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_{2}\mathrm{O}\left(aq\right)}\phantom{\rule{0ex}{0ex}}$

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