Balancing equations
a) Mg + HCl --> MgCl2 + H2
b) K + H2O --> KOH + H2
c) Li + H2O --> LiOH + H2
d) Fe + O2 --> Fe2O3
Please help
To balance chemical equations, you need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how you can balance each of the given equations:
a) Mg + HCl → MgCl2 + H2
- Let's start by counting the number of atoms on each side.
Mg: 1 on the left side, 1 on the right side
H: 1 on the left side, 2 on the right side
Cl: 1 on the left side, 2 on the right side
- To balance the hydrogen atoms, we need to put a coefficient of 2 in front of HCl:
Mg + 2HCl → MgCl2 + H2
- Now we have 2 chlorine atoms on the right side, so we put a coefficient of 2 in front of MgCl2:
Mg + 2HCl → 2MgCl2 + H2
- Finally, we have 2 magnesium atoms on the right side, so we put a coefficient of 2 in front of Mg:
2Mg + 2HCl → 2MgCl2 + H2
b) K + H2O → KOH + H2
- Let's count the number of atoms:
K: 1 on the left side, 1 on the right side
H: 2 on the left side, 2 on the right side
O: 1 on the left side, 1 on the right side
- The equation is already balanced. There are equal numbers of atoms on both sides of the equation:
K + H2O → KOH + H2
c) Li + H2O → LiOH + H2
- Atom counts:
Li: 1 on the left side, 1 on the right side
H: 2 on the left side, 2 on the right side
O: 1 on the right side, 1 in the LiOH molecule on the right side
- To balance the oxygen atoms, we need to put a coefficient of 2 in front of LiOH:
Li + H2O → 2LiOH + H2
- Now, the equation is balanced with the same number of atoms on both sides.
d) Fe + O2 → Fe2O3
- Counting the atoms:
Fe: 1 on the left side, 2 on the right side
O: 2 on the left side, 3 on the right side
- To balance the iron atoms, put a coefficient of 2 in front of Fe on the left side:
2Fe + O2 → Fe2O3
- Now, both sides have 2 iron atoms.
- Balancing the oxygen atoms, put a coefficient of 3/2 in front of O2 on the left side:
2Fe + (3/2)O2 → Fe2O3
- To avoid fractions, multiply all coefficients by 2:
4Fe + 3O2 → 2Fe2O3
- The equation is now balanced.