# bacl2 + h2so4

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## Balancing a chemical reaction:Chemical equations are symbolic representations of chemical reactions where the reactants and products are expressed in terms of their respective chemical formulae.To balance a chemical reaction we first need to tướng write an unbalanced equation of the reaction.Then we have to tướng count the number of atoms except for hydrogen and oxygen present on the left-hand side and on the ride-side.Then we have to tướng write the coefficient in front of that compound.At last balancing the hydrogen and oxygen atoms because they are present in more than vãn one compound.The reaction of Barium chloride and sulfuric acid:Barium chloride reacts with sulfuric acid to tướng size barium sulfate and hydrochloric acid.The unbalanced equation for this is $\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_{2}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\left(\mathrm{l}\right)}\to \underset{\mathrm{barium}\mathrm{sulfate}}{{\mathrm{BaSO}}_{4}\left(\mathrm{s}\right)}+\underset{\mathrm{hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{l}\right)}$Balancing the number of chlorine atoms on both sides by adding $2$ coefficient before $\mathrm{HCl}$ we get, $\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_{2}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\left(\mathrm{l}\right)}\to \underset{\mathrm{barium}\mathrm{sulfate}}{{\mathrm{BaSO}}_{4}\left(\mathrm{s}\right)}+\underset{\mathrm{hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{l}\right)}$The Barium, Sulfur, Hydrogen, and Oxygen atoms are automatically balanced.Therefore, the balanced equation is $\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_{2}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\left(\mathrm{l}\right)}\to \underset{\mathrm{barium}\mathrm{sulfate}}{{\mathrm{BaSO}}_{4}\left(\mathrm{s}\right)}+\underset{\mathrm{hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{l}\right)}$

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