How do I balance the equations :Ca2CO3+ HCl2=CaCl+CO2+H2O

Why do people call these "word equations"?

The first thing to do is to correctly write down the chemical formulas. You can't balance it unless they're correct.

CaCO_3 + HCl \rightleftharpoons CaCl_2 + CO_2 + H_2O

This actually occurs in two steps. If you understand that, it makes it easier. It's still possible to balance it if you don't.

CaCO_3 + HCl \rightleftharpoons CaCl_2 + H_2CO_3 \rightleftharpoons CaCl_2 + CO_2 + H_2O

It's easiest to understand that some groups of elements act as a single group. For example, carbonate acts as one group (CO3^(-2)). This is easily seen if you write down the ionic form

CaCO_3 \rightleftharpoons Ca^{+2} + CO_3^{-2}

HCl \rightleftharpoons H^+ + Cl^-

Proceed from left to right, treating the grouped anions together. First we try to balance the calcium -- it's already balanced. Then we try to balance the chloride since calcium goes to calcium chloride. We need two chlorine atoms so we fix the equation appropriately (2 HCl).

CaCO_3 + 2HCl \rightleftharpoons CaCl_2 + CO_2 + H_2O

We check the carbons -- they're balanced. If they weren't, we might need to change the number of CaCO3 -- and also the number of CaCl2. In this case, we don't need to change anything. We check the hydrogens and double-check the chlorines. Viola, it's balanced.

how do I balance the equations:Pb(NO3)4 +Na2CO3+H2o

I meant Pb(NO3)4+Na2SO4=Pb(SO4)4+NaNO3

Count the number of elements on one side and try making the numbers equal on both sides by putting a number in front of the formula, like Perito said.

And you did another mistake there, Pb has valency 2 or 4, but not higher, so, Pb(SO4)2 instead of Pb(SO4)4.

Pb(NO_3)_4+Na_2SO_4=Pb(SO_4)_2+NaNO_3

Now that is easier since you can see some of the elements as groups. Earlier you were taught with CO3, now, we'll talk about NO3 and SO4.

So that there is a same amount of NO3 on both sides, put '4' in front of the NaNO3.

For the SO4, put '2' in front of the Na2SO4.


Pb(NO_3)_4+2Na_2SO_4=Pb(SO_4)_2+4NaNO_3