what is meant by skeletal equation ?why it should be balanced? balance some of the difficult skeletal equation?
what is meant by skeletal equation ?why it should be balanced? balance some of the difficult skeletal equation?
The chemical equations which is not balanced and does not tell us anything about the reactions for example it's type,the state of the reactants and the catalyst used is known as a skeletal equation There are a couple of reasons why you have to balance equations.
Firstly, because it tells you the relative quantities of re-agents you need, and the relative quantities of products you’ll get.
Secondly, if you want to calculate how much energy is absorbed/released in the reaction, you need to know exactly how many of each chemical in the formula is required.
To balance the equation, I write out the equation, and then count the number of instances of each element on each side of the “=” sign. Then, for each element, compare the number from the LHS & RHS of the equation. Where there’s a difference, you have to start putting multipliers in front of the atoms/molecules in order to increase one side. Of course, this may then change the numbers for other elements, so you have to keep recalculating, and adding multipliers until everything balances. There may be a more formal way of doing it, but I use intelligent guesswork!
I’ll give an example covering both parts of your question. I’ll use the equation for reacting hydrogen and oxygen to make water.
H2+O2=H2OH2+O2=H2O
That looks right doesn’t it? Well, lets count the atoms. Hydrogen: 2 on the left and 2 on the right - that’s balanced. Oxygen: 2 on the left and 1 on the right. That’s not balanced - we need more on the right. Let’s multiple the H2OH2O by 2:
H2+O2=2H2OH2+O2=2H2O
Okay, now the oxygens have 2 atoms on each side, which is good, but the hydrogens have 2 on the left and 4 on the right. We need to increase the hydrogens on the left:
2H2+O2=2H2O2H2+O2=2H2O
Okay, lets count again: 4 hydrogens on the left and 4 on the right; 2 oxygens on the left and 2 on the right. Everything balances, and so we’re finished.
Now what does it tell us? It tells us that we need twice as much hydrogen as oxygen if we want all the re-agents to react to form water. I’m sure you know what a mole is in chemistry. If you tried to take 1 mole of hydrogen and 1 mole of oxygen, mixed them together and set light to it, you would have half of the oxygen left over at the end. So you see why it’s important to know the relative quantities of the re-agents.
Problem #1: FeCl3 + MgO ---> Fe2O3 + MgCl2
Solution:
1) Balance the Cl (note that 2 x 3 = 3 x 2):
2FeCl3 + MgO ---> Fe2O3 + 3MgCl2 The Fe also gets balanced in this step.
2) Pick either the O or the Mg to balance next:
2FeCl3 + 3MgO ---> Fe2O3 + 3MgCl2 The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
Problem #2: Li + H3PO4 ---> H2 + Li3PO4
Solution:
1) Balance the Li:
3Li + H3PO4 ---> H2 + Li3PO4 2) Now, look at the hydrogens. See how the H comes only in groups of 3 on the left and only in groups of 2 on the right? Do this:
3Li + 2H3PO4 ---> 3H2 + Li3PO4 Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3) Balance the phosphate as a group:
3Li + 2H3PO4 ---> 3H2 + 2Li3PO4 4) Oops, that messed up the lithium, so we fix it:
6Li + 2H3PO4 ---> 3H2 + 2Li3PO4 Problem #3: ZnS + O2 ---> ZnO + SO2
1) Balance the oxygen with a fractional coefficient (Zn and S are already balanced):
ZnS + (3/2)O2 ---> ZnO + SO2 2) Multiply through to clear the fraction:
2ZnS + 3O2 ---> 2ZnO + 2SO2
There are a couple of reasons why you have to balance equations.
Firstly, because it tells you the relative quantities of re-agents you need, and the relative quantities of products you’ll get.
Secondly, if you want to calculate how much energy is absorbed/released in the reaction, you need to know exactly how many of each chemical in the formula is required.
To balance the equation, I write out the equation, and then count the number of instances of each element on each side of the “=” sign. Then, for each element, compare the number from the LHS & RHS of the equation. Where there’s a difference, you have to start putting multipliers in front of the atoms/molecules in order to increase one side. Of course, this may then change the numbers for other elements, so you have to keep recalculating, and adding multipliers until everything balances. There may be a more formal way of doing it, but I use intelligent guesswork!
I’ll give an example covering both parts of your question. I’ll use the equation for reacting hydrogen and oxygen to make water.
H2+O2=H2OH2+O2=H2O
That looks right doesn’t it? Well, lets count the atoms. Hydrogen: 2 on the left and 2 on the right - that’s balanced. Oxygen: 2 on the left and 1 on the right. That’s not balanced - we need more on the right. Let’s multiple the H2OH2O by 2:
H2+O2=2H2OH2+O2=2H2O
Okay, now the oxygens have 2 atoms on each side, which is good, but the hydrogens have 2 on the left and 4 on the right. We need to increase the hydrogens on the left:
2H2+O2=2H2O2H2+O2=2H2O
Okay, lets count again: 4 hydrogens on the left and 4 on the right; 2 oxygens on the left and 2 on the right. Everything balances, and so we’re finished.
Now what does it tell us? It tells us that we need twice as much hydrogen as oxygen if we want all the re-agents to react to form water. I’m sure you know what a mole is in chemistry. If you tried to take 1 mole of hydrogen and 1 mole of oxygen, mixed them together and set light to it, you would have half of the oxygen left over at the end. So you see why it’s important to know the relative quantities of the re-agents.
Problem #1: FeCl3 + MgO ---> Fe2O3 + MgCl2
Solution:
1) Balance the Cl (note that 2 x 3 = 3 x 2):
2FeCl3 + MgO ---> Fe2O3 + 3MgCl2The Fe also gets balanced in this step.
2) Pick either the O or the Mg to balance next:
2FeCl3 + 3MgO ---> Fe2O3 + 3MgCl2The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
Problem #2: Li + H3PO4 ---> H2 + Li3PO4
Solution:
1) Balance the Li:
3Li + H3PO4 ---> H2 + Li3PO42) Now, look at the hydrogens. See how the H comes only in groups of 3 on the left and only in groups of 2 on the right? Do this:
3Li + 2H3PO4 ---> 3H2 + Li3PO4Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3) Balance the phosphate as a group:
3Li + 2H3PO4 ---> 3H2 + 2Li3PO44) Oops, that messed up the lithium, so we fix it:
6Li + 2H3PO4 ---> 3H2 + 2Li3PO4Problem #3: ZnS + O2 ---> ZnO + SO2
1) Balance the oxygen with a fractional coefficient (Zn and S are already balanced):
ZnS + (3/2)O2 ---> ZnO + SO22) Multiply through to clear the fraction:
2ZnS + 3O2 ---> 2ZnO + 2SO2
