Answer
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Hint:. We can write the chemical equation of the above given chemical reactions, then by applying trial and error methods in order to balance the chemical equation. The condition to balance is the number of a particular element in the reactant should be equal to the number of elements in the product.
Complete step by step answer:
Chemical equation is the representation of a chemical reaction where the reactants react to give products. The law of conservation of mass attributes for balancing a chemical equation. The law of conservation of mass states that the atoms are neither created nor destroyed. In a chemical equation the substance that is left to the arrow is called reactant and the substance that is to the right of the arrow is called product. Applying the law of conservation of mass to a chemical, there is no atom destroyed from the reactant side nor created in the product, so the number of elements in the reactant is equal to the number of elements in the product. This is the theory behind balancing chemical equations.
(a) Hydrogen gas combines with nitrogen to form ammonia.
Chemical equation: ${{H}_{2}}+{{N}_{2}}\overset{{}}{\leftrightarrow}N{{H}_{3}}$
Now, let's balance the equation:
- The number of H atoms in the reactant is 2 but, on the product side the number of H atoms is 3. So, in order to balance this equation we should put 3 before the hydrogen in the reactant and add 2 to ammonia in the product.
\[3{{H}_{2}}+{{N}_{2}}\overset{{}}{\leftrightarrow}2N{{H}_{3}}\]
Now, all elements are equal in number on both sides. Thus, this is the balanced chemical equation.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
Chemical equation: ${{H}_{2}}S+{{O}_{2}}\xrightarrow{{}}{{H}_{2}}O+S{{O}_{2}}$
Now, let's balance the equation:
-The number of H, S atoms is equal on both sides but, O atom on both sides are not equal.
So, let's add 3 to oxygen in the reactant side then, now there are 6 atoms in the reactant. We can add 2 before water and sulphur dioxide in the product. ${{H}_{2}}S+3{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+2S{{O}_{2}}$
Now H atom and S atom are unequal so, we can add 2 in front of hydrogen sulphide. \[2{{H}_{2}}S+3{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+2S{{O}_{2}}\]
Thus, this is the balanced chemical equation for this reaction.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
Chemical equation: \[BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}BaS{{O}_{4}}+AlC{{l}_{3}}\]
Number of Cl atom in reactant is 2 but the Cl atom in product is 3. so in order to balance it we should add 3 before $BaC{{l}_{2}}$ and 2 before $AlC{{l}_{3}}$.
\[3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}BaS{{O}_{4}}+2AlC{{l}_{3}}\]
Now, we can see that the Ba atom is unequal, so we should add 2 in front of barium sulphate in product.
Thus, the balanced equation is:
\[3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}3BaS{{O}_{4}}+2AlC{{l}_{3}}\]
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Chemical equation: \[K+{{H}_{2}}O\xrightarrow[{}]{}KOH+{{H}_{2}}\]
The number of hydrogen atoms is 2 in the reactant side but there are 3 hydrogen atoms on the product side. So, we should add 2 before K, ${{H}_{2}}O$ and KOH .
\[2K+2{{H}_{2}}O\xrightarrow[{}]{}2KOH+{{H}_{2}}\]
Thus, this is the balanced equation.
Note: When we are taking a number of elements, if the element has both subscript and coefficient in order to get a number of elements we should multiply subscript and coefficient. Only one simple principle is that the number of a particular element in the reactant should be equal to the number of elements in the product.
Complete step by step answer:
Chemical equation is the representation of a chemical reaction where the reactants react to give products. The law of conservation of mass attributes for balancing a chemical equation. The law of conservation of mass states that the atoms are neither created nor destroyed. In a chemical equation the substance that is left to the arrow is called reactant and the substance that is to the right of the arrow is called product. Applying the law of conservation of mass to a chemical, there is no atom destroyed from the reactant side nor created in the product, so the number of elements in the reactant is equal to the number of elements in the product. This is the theory behind balancing chemical equations.
(a) Hydrogen gas combines with nitrogen to form ammonia.
Chemical equation: ${{H}_{2}}+{{N}_{2}}\overset{{}}{\leftrightarrow}N{{H}_{3}}$
Now, let's balance the equation:
- The number of H atoms in the reactant is 2 but, on the product side the number of H atoms is 3. So, in order to balance this equation we should put 3 before the hydrogen in the reactant and add 2 to ammonia in the product.
\[3{{H}_{2}}+{{N}_{2}}\overset{{}}{\leftrightarrow}2N{{H}_{3}}\]
Now, all elements are equal in number on both sides. Thus, this is the balanced chemical equation.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
Chemical equation: ${{H}_{2}}S+{{O}_{2}}\xrightarrow{{}}{{H}_{2}}O+S{{O}_{2}}$
Now, let's balance the equation:
-The number of H, S atoms is equal on both sides but, O atom on both sides are not equal.
So, let's add 3 to oxygen in the reactant side then, now there are 6 atoms in the reactant. We can add 2 before water and sulphur dioxide in the product. ${{H}_{2}}S+3{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+2S{{O}_{2}}$
Now H atom and S atom are unequal so, we can add 2 in front of hydrogen sulphide. \[2{{H}_{2}}S+3{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+2S{{O}_{2}}\]
Thus, this is the balanced chemical equation for this reaction.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
Chemical equation: \[BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}BaS{{O}_{4}}+AlC{{l}_{3}}\]
Number of Cl atom in reactant is 2 but the Cl atom in product is 3. so in order to balance it we should add 3 before $BaC{{l}_{2}}$ and 2 before $AlC{{l}_{3}}$.
\[3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}BaS{{O}_{4}}+2AlC{{l}_{3}}\]
Now, we can see that the Ba atom is unequal, so we should add 2 in front of barium sulphate in product.
Thus, the balanced equation is:
\[3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\xrightarrow{{}}3BaS{{O}_{4}}+2AlC{{l}_{3}}\]
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Chemical equation: \[K+{{H}_{2}}O\xrightarrow[{}]{}KOH+{{H}_{2}}\]
The number of hydrogen atoms is 2 in the reactant side but there are 3 hydrogen atoms on the product side. So, we should add 2 before K, ${{H}_{2}}O$ and KOH .
\[2K+2{{H}_{2}}O\xrightarrow[{}]{}2KOH+{{H}_{2}}\]
Thus, this is the balanced equation.
Note: When we are taking a number of elements, if the element has both subscript and coefficient in order to get a number of elements we should multiply subscript and coefficient. Only one simple principle is that the number of a particular element in the reactant should be equal to the number of elements in the product.
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