46. Balance these equations:
(a) ____Fe(s) + ____O2(g)h____ Fe2O3(s)
(b) ____H2(g) + ____N2(g)h____NH3(g)
(c) ____Cl2(g) + ____KBr(aq)h____Br2(l ) +
____KCl (aq)
(d) ____CH4(g) + ____O2(g)h____CO2(g) +
____H2O(l )
What does the symbol h represent?
The arrow from reactants to products?
Here is one answer.
3 H2(g) + N2(g)-> 2 NH3(g)
The others are not hard. You need to try them yourself.
To balance these chemical equations, we need to ensure that the same number of atoms of each element are present on both sides of the equation. We can achieve this by adjusting the coefficients in front of each reactant and product.
(a) ____Fe(s) + ____O2(g) ⟶ ____Fe2O3(s)
To balance the equation for the reaction between iron and oxygen gas to form iron(III) oxide, Fe2O3, we need to make sure that the number of Fe and O atoms is the same on both sides.
Counting the atoms, we have:
Fe: 1 on the left and 2 on the right
O: 2 on the left and 3 on the right
We can start by placing a coefficient of 2 in front of Fe on the left side:
2 Fe(s) + ____O2(g) ⟶ ____Fe2O3(s)
This now gives us:
Fe: 2 on the left and 2 on the right
O: 2 on the left and 3 on the right
To balance the oxygen atoms, we need to place a coefficient of 3/2 or 1.5 in front of O2 on the left side:
2 Fe(s) + 1.5 O2(g) ⟶ ____Fe2O3(s)
Lastly, to make sure all the coefficients are whole numbers, we can multiply the entire equation by 2:
4 Fe(s) + 3 O2(g) ⟶ 2 Fe2O3(s)
So, balanced equation (a) is:
4 Fe(s) + 3 O2(g) ⟶ 2 Fe2O3(s)
You can apply a similar approach to balance equations (b), (c), and (d).
(b) ____H2(g) + ____N2(g) ⟶ ____NH3(g)
(c) ____Cl2(g) + ____KBr(aq) ⟶ ____Br2(l) + ____KCl(aq)
(d) ____CH4(g) + ____O2(g) ⟶ ____CO2(g) + ____H2O(l)