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# Translate the following statements into chemical equations and then balance them:(i) Hydrogen gas combines with nitrogen to give ammonia.(ii) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.(iii) Barium chloride reacts with ammonium sulphate to give aluminium chloride and a precipitate of barium sulphate.(iv) Potassium metal reacts with water to give potassium hydroxide and hydrogen

Last updated date: 13th Jun 2024
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Hint: While balancing any equation, one needs to multiply the chemical species by such numbers that all the atoms taking part in the reaction are present in the same numbers at each side( reactant side and product side) of the reaction.

Complete step by step answer:
According to the law of conservation of mass, the total mass of the products needs to be equal to the total mass of reactants. In other words the number of any atoms on the left side of the equation have to be equal to the number of that atom present in the right side of the equation. If this is not followed for any equation, then the equation is said to be unbalanced.
To balance any equation, we just need to make sure that the same number of particular atoms are present on both sides of the equation and if not then just multiply any of the species by the required number and make them balanced.
Let’s see balancing of equations given in the options.
(i) Hydrogen gas combines with nitrogen to give ammonia.
We can write this in the form of reaction as ${H_{2(g)}} + {N_{2(g)}} \to N{H_{3(g)}}$ but here number of nitrogen and hydrogen atoms on right hand side and left side are not equal. To make them equal we need to multiply them. If we multiply ${H_{2(g)}}$ by 3 then we will get 2 moles of $N{H_{3(g)}}$ as a product and hence the equation will be balanced.
Balanced equation: $3{H_{2(g)}} + {N_{2(g)}} \to 2N{H_{3(g)}}$
(ii) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
we can write the unbalanced reaction as ${H_2}{S_{(g)}} + {O_{2(g)}} \to {H_2}{O_{(l)}} + S{O_{2(g)}}$
Now, see that number of oxygen atoms on both sides are not equal. So, we will need to balance this. 2 hydrogen atoms are there in only one species on both of the sides, so we can say that the number that both hydrogen containing species are multiplied with, need to be equal. Only way we can balance this equation is by multiplying ${H_2}{S_{(g)}}$ by 2, ${O_{2(g)}}$ by 3, ${H_2}{O_{(l)}}$ by 2 and $S{O_{2(g)}}$ by 2.
So, balanced equation will be, $2{H_2}{S_{(g)}} + 3{O_{2(g)}} \to 2{H_2}{O_{(l)}} + 2S{O_{2(g)}}$
(iii) Barium chloride reacts with ammonium sulphate to give aluminium chloride and a precipitate of barium sulphate.
Unbalanced equation for this reaction will be, $BaC{l_{2(aq)}} + {(N{H_4})_2}S{O_{4(aq)}} \to N{H_4}C{l_{(aq)}} + BaS{O_{4(s)}}$
Now, if we just multiply $N{H_4}C{l_{(aq)}}$ by 2, then we will get the balanced equation.
So, balanced equation for this reaction will be $BaC{l_{2(aq)}} + {(N{H_4})_2}S{O_{4(aq)}} \to 2N{H_4}C{l_{(aq)}} + BaS{O_{4(s)}}$
(iv) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Unbalanced equation can be given as ${K_{(s)}} + {H_2}{O_{(l)}} \to KO{H_{(aq)}} + {H_{2(g)}}$
We can see that the number we multiply with ${K_{(s)}}$ and $KO{H_{(aq)}}$ needs to be same because they are only Potassium containing species on their respective side. But the number of hydrogens are not balanced. So, we will need to multiply ${K_{(s)}}$, $KO{H_{(aq)}}$ and ${H_2}{O_{(l)}}$ with 2 in order to get the balanced equation.
So, the balanced equation will be $2{K_{(s)}} + 2{H_2}{O_{(l)}} \to 2KO{H_{(aq)}} + {H_{2(g)}}$.

The process of balancing the equation is purely arbitrary as we just need to assign different numbers ahead of species taking part in the reaction and in the end we need to assign numbers in such a way that numbers of each atom on each side is equal.

Note:
Do not forget to re-check the numbers of atoms present on each side as sometimes when in a hurry, balancing of equations may have errors. Note that all the numbers we assign to species need to be as small as possible as an example $2{K_{(s)}} + 2{H_2}{O_{(l)}} \to 2KO{H_{(aq)}} + {H_{2(g)}}$.....(1)
$4{K_{(s)}} + 4{H_2}{O_{(l)}} \to 4KO{H_{(aq)}} + 2{H_{2(g)}}$....(2)
Here both reactions are balanced but (1) is more proper because it has the least number of reactants. So, if all the numbers are divisible by the same number, the dividing of all the numbers assigned needs to be done.