balanced equation

mg+ pbcl2---mgcl2+pb
F+NaBr--FBr+Na

Your laziness leaves me bewildered. I suppose we are supposed to know mg is magnesium and not milligrams. And mgcl2 (whew!).

Mg + PbCl2 ==> MgCl2 + Pb
You should be able to count and know this equation is balanced as is.

F + NaBr = FBr + Na. What comic book did you use to find this one. Again, it is balanced as is and you should be able to count atoms and know that it is balanced. BUT the products are not right. You should have
F2 + 2NaBr ==> 2NaF + Br2

thank you.

I always want to know if i am doing the right procedure for calculations.so i want make your if i did balanced the equation.

To balance the chemical equations provided, follow these steps:

1. Write down the unbalanced equation:
a) Mg + PbCl2 → MgCl2 + Pb
b) F + NaBr → FBr + Na

2. Count the number of atoms for each element on both sides of the equation.

For equation (a):
Mg: 1 on the left, 1 on the right
Pb: 1 on the left, 1 on the right
Cl: 2 on the left, 2 on the right

For equation (b):
F: 1 on the left, 1 on the right
Na: 1 on the left, 1 on the right
Br: 1 on the left, 1 on the right

3. Start by balancing the atoms of elements that appear in only one compound on each side of the equation.

For equation (a):
The magnesium (Mg) atoms are already balanced.
The lead (Pb) atoms are already balanced.
The chlorine (Cl) atoms are already balanced.

For equation (b):
The fluorine (F) atoms are already balanced.
The sodium (Na) atoms are already balanced.
The bromine (Br) atoms are already balanced.

4. Next, balance the atoms of elements that appear in multiple compounds, one at a time.

For equation (a):
Since there are two Cl atoms on the right side, we need to balance the Cl atoms on the left side. Therefore, the balanced equation is:
Mg + PbCl2 → MgCl2 + Pb

For equation (b):
Since there is one Br atom on the left side, we need to balance the Br atoms on the right side. Therefore, the balanced equation is:
F + NaBr → FBr + Na

The balanced equations are:
a) Mg + PbCl2 → MgCl2 + Pb
b) F + NaBr → FBr + Na