
Balance the following equations :
a.) $CaC{O_3} + HCl \to CaC{l_2} + C{O_{_2}} \uparrow + {H_2}O$
b.) $Na + {H_2}O \to NaOH + {H_2} \uparrow $
c.) ${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + N{H_3} + {H_2}O$
d.) $Hg{(N{O_3})_2} \to Hg + N{O_2} \uparrow + {O_2} \uparrow $
e.) $P{b_3}{O_4} + HCl \to PbC{l_2} + C{l_2} + {H_2}O$
Answer
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Hint: By using hit and trial method, we can easily balance the equations. In such a method, one needs to find the number of atoms of each type of element and then balance them by multiplying to a certain number. The method is named so because there is not any formula that will tell us by which number one has to multiply the element. It is always based on our assumptions.
Complete step by step solution:
A balanced chemical equation is the one in which the number of atoms for each element in the reaction and total charge is the same for both reactants and products. There are some steps to be followed to balance the equations.
Step 1 :- Identify the reactants and products.
Step 2 :- Write an equation for the reaction.
Step 3 :- Count the number of atoms of each element in the reactants and products.
Step 4 :- Balance the equation.
So, the reactions can be balanced as-
$CaC{O_3} + HCl \to CaC{l_2} + C{O_{_2}} \uparrow + {H_2}O$
Now is the step to count the number of atoms of each element. These are –
From the table we see that Ca, C and O are equal. So, no changes.
But H and Cl are 2 in number on the product side. Thus, in reactants they can be multiplied by 2 i.e. we have 2 molecules of HCl.
So, the new balanced equation is :- $CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} \uparrow + {H_2}O$
Second equation is :-
$Na + {H_2}O \to NaOH + {H_2} \uparrow $
Now count the number of atoms of each element. These are –
From the table we see that Na are equal. So, no changes.
On the observing table; we find that by multiplying Na, NaOH and ${H_2}O$ by 2; we get an equal number of atoms of all elements on both sides.
Thus, the balanced chemical equation is :
$2Na + 2{H_2}O \to 2NaOH + {H_2} \uparrow $
Third equation is :-
${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + N{H_3} + {H_2}O$
First, we will count the number of atoms as –
On observing the atoms we find out that by multiplying 2 with ammonia and water on product side; we get balanced equation which is –
${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + 2N{H_3} + 2{H_2}O$
Fourth equation is :-
$Hg{(N{O_3})_2} \to Hg + N{O_2} \uparrow + {O_2} \uparrow $
Again, we will first count the number of atoms of each element.
From the table, we observe that to balance the equation; we need two N atoms on the product side which is one and 6 oxygen atoms on the product side which is 4.
Thus, multiplying 2 with nitrogen dioxide on the product side will give us a balanced equation.
Now, the last and fifth equation is :-
$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O$$P{b_3}{O_4} + HCl \to PbC{l_2} + C{l_2} + {H_2}O$
So, let’s first count the number of atoms.
On observing the table, we find that by multiplying HCl with 8 on reactant side, $PbC{l_2}$with 3 and water with 4 on product side; we get the balanced equation as –$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O$
Note:
Balancing the equations is easy if you count the number of atoms first. Always try to make a table during comparison of atoms on reactant and product sides. Further, if in any case we have some element which is equal on both sides but that element is in molecular form with other atoms and that atoms are needed to be balanced. Then, after balancing the second element we should check that the element which was earlier equal should now also be equal on both sides.
Example – In the first equation, sodium was earlier equal in reactant and product side but after balancing other elements, it was again needed to be balanced.
Complete step by step solution:
A balanced chemical equation is the one in which the number of atoms for each element in the reaction and total charge is the same for both reactants and products. There are some steps to be followed to balance the equations.
Step 1 :- Identify the reactants and products.
Step 2 :- Write an equation for the reaction.
Step 3 :- Count the number of atoms of each element in the reactants and products.
Step 4 :- Balance the equation.
So, the reactions can be balanced as-
$CaC{O_3} + HCl \to CaC{l_2} + C{O_{_2}} \uparrow + {H_2}O$
Now is the step to count the number of atoms of each element. These are –
| Element | Reactant | Product |
| Ca | 1 | 1 |
| C | 1 | 1 |
| O | 3 | 3 |
| H | 1 | 2 |
| Cl | 1 | 2 |
From the table we see that Ca, C and O are equal. So, no changes.
But H and Cl are 2 in number on the product side. Thus, in reactants they can be multiplied by 2 i.e. we have 2 molecules of HCl.
So, the new balanced equation is :- $CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} \uparrow + {H_2}O$
Second equation is :-
$Na + {H_2}O \to NaOH + {H_2} \uparrow $
Now count the number of atoms of each element. These are –
| Element | Reactant | Product |
| Na | 1 | 1 |
| H | 2 | 3 |
| O | 1 | 1 |
From the table we see that Na are equal. So, no changes.
On the observing table; we find that by multiplying Na, NaOH and ${H_2}O$ by 2; we get an equal number of atoms of all elements on both sides.
Thus, the balanced chemical equation is :
$2Na + 2{H_2}O \to 2NaOH + {H_2} \uparrow $
Third equation is :-
${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + N{H_3} + {H_2}O$
First, we will count the number of atoms as –
| Element | Reactant | Product |
| N | 2 | 1 |
| H | 10 | 5 |
| S | 1 | 1 |
| O | 2 | 1 |
| Ca | 1 | 1 |
| $S{O_4}$ | 1 | 1 |
On observing the atoms we find out that by multiplying 2 with ammonia and water on product side; we get balanced equation which is –
${(N{H_4})_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + 2N{H_3} + 2{H_2}O$
Fourth equation is :-
$Hg{(N{O_3})_2} \to Hg + N{O_2} \uparrow + {O_2} \uparrow $
Again, we will first count the number of atoms of each element.
| Element | Reactant | Product |
| Hg | 1 | 1 |
| N | 2 | 1 |
| O | 6 | 4 |
From the table, we observe that to balance the equation; we need two N atoms on the product side which is one and 6 oxygen atoms on the product side which is 4.
Thus, multiplying 2 with nitrogen dioxide on the product side will give us a balanced equation.
Now, the last and fifth equation is :-
$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O$$P{b_3}{O_4} + HCl \to PbC{l_2} + C{l_2} + {H_2}O$
So, let’s first count the number of atoms.
| Element | Reactant | Product |
| Pb | 3 | 1 |
| O | 4 | 1 |
| H | 1 | 2 |
| Cl | 1 | 4 |
On observing the table, we find that by multiplying HCl with 8 on reactant side, $PbC{l_2}$with 3 and water with 4 on product side; we get the balanced equation as –$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + C{l_2} + 4{H_2}O$
Note:
Balancing the equations is easy if you count the number of atoms first. Always try to make a table during comparison of atoms on reactant and product sides. Further, if in any case we have some element which is equal on both sides but that element is in molecular form with other atoms and that atoms are needed to be balanced. Then, after balancing the second element we should check that the element which was earlier equal should now also be equal on both sides.
Example – In the first equation, sodium was earlier equal in reactant and product side but after balancing other elements, it was again needed to be balanced.
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