Answered

Balance each chemical equation and identify the type of reaction:



1) $\_ N_2 + \_ H_2 \longrightarrow \_ NH_3$

2) $\_ Na + \_ H_2O \longrightarrow \_ NaOH + \_ H_2$

3) $\_ KClO_3 \longrightarrow \_ KCl + \_ O_2$

4) $\_ NaCl + \_ F_2 \longrightarrow \_ NaF + \_ Cl_2$

5) $\_ H_2 + \_ O_2 \longrightarrow \_ H_2O$

6) $\_ Pb(OH)_2 + \_ HCl \longrightarrow \_ H_2O + \_ PbCl_2$

7) $\_ AlBr_3 + \_ K_2SO_4 \longrightarrow \_ KBr + \_ Al_2(SO_4)_3$

8) $\_ C_3H_8 + \_ O_2 \longrightarrow \_ CO_2 + \_ H_2O$

9) $\_ FeCl_3 + \_ NaOH \longrightarrow \_ Fe(OH)_3 + \_ NaCl$

Verified Answer :

  • Balance each reaction by applying the Law of Conservation of Mass, ensuring equal atom counts for all elements.
  • The balanced coefficients are:

1) \(1 N_2 + 3 H_2 \longrightarrow 2 NH_3\)
2) \(2 Na + 2 H_2O \longrightarrow 2 NaOH + 1 H_2\)
3) \(2 KClO_3 \longrightarrow 2 KCl + 3 O_2\)
4) \(2 NaCl + 1 F_2 \longrightarrow 2 NaF + 1 Cl_2\)
5) \(2 H_2 + 1 O_2 \longrightarrow 2 H_2O\)
6) \(1 Pb(OH)_2 + 2 HCl \longrightarrow 2 H_2O + 1 PbCl_2\)
7) \(2 AlBr_3 + 3 K_2SO_4 \longrightarrow 6 KBr + 1 Al_2(SO_4)_3\)
8) \(1 C_3H_8 + 5 O_2 \longrightarrow 3 CO_2 + 4 H_2O\)
9) \(1 FeCl_3 + 3 NaOH \longrightarrow 1 Fe(OH)_3 + 3 NaCl\)

  • Final results are verified for atom conservation per element.
  • \(\boxed{1N_2+3H_2\rightarrow 2NH_3, 2Na+2H_2O\rightarrow 2NaOH+H_2, 2KClO_3\rightarrow 2KCl+3O_2, 2NaCl+F_2\rightarrow 2NaF+Cl_2, 2H_2+O_2\rightarrow 2H_2O, Pb(OH)_2+2HCl\rightarrow 2H_2O+PbCl_2, 2AlBr_3+3K_2SO_4\rightarrow 6KBr+Al_2(SO_4)_3, C_3H_8+5O_2\rightarrow 3CO_2+4H_2O, FeCl_3+3NaOH\rightarrow Fe(OH)_3+3NaCl}\)

Explanation

  1. Analysis of the Problem

To balance chemical equations, we apply the Law of Conservation of Mass, which states that the number of atoms of each element must remain the same on both the reactant and product sides. We assign coefficients to each compound and equate the atom counts for each element involved in the reaction. If an element appears in more than one compound, we prioritize the balancing of unique elements first to simplify the algebraic system.

  1. Balancing Reactions 1 through 3

1) \(\underline{1} N_2 + \underline{3} H_2 \longrightarrow \underline{2} NH_3\)
Balancing N: 2 on left, 2 on right (via 2 \(NH_3\)).
Balancing H: 6 on right (23), so need 3 \(H_2\) on left.
Result: \(1 N_2 + 3 H_2 \longrightarrow 2 NH_3\)

2) \(\underline{2} Na + \underline{2} H_2O \longrightarrow \underline{2} NaOH + \underline{1} H_2\)
Balancing Na: 2 on left, 2 on right.
Balancing O: 2 on left, 2 on right.
Balancing H: 4 on left (22), 4 on right (2+2).
Result: \(2 Na + 2 H_2O \longrightarrow 2 NaOH + 1 H_2\)

3) \(\underline{2} KClO_3 \longrightarrow \underline{2} KCl + \underline{3} O_2\)
Balancing K and Cl: 2 on each side.
Balancing O: 6 on left (23), 6 on right (32).
Result: \(2 KClO_3 \longrightarrow 2 KCl + 3 O_2\)

  1. Balancing Reactions 4 through 6

4) \(\underline{2} NaCl + \underline{1} F_2 \longrightarrow \underline{2} NaF + \underline{1} Cl_2\)
Balancing Na: 2 on left, 2 on right.
Balancing Cl: 2 on left, 2 on right.
Balancing F: 2 on left, 2 on right.
Result: \(2 NaCl + 1 F_2 \longrightarrow 2 NaF + 1 Cl_2\)

5) \(\underline{2} H_2 + \underline{1} O_2 \longrightarrow \underline{2} H_2O\)
Balancing H: 4 on left, 4 on right.
Balancing O: 2 on left, 2 on right.
Result: \(2 H_2 + 1 O_2 \longrightarrow 2 H_2O\)

6) \(\underline{1} Pb(OH)_2 + \underline{2} HCl \longrightarrow \underline{2} H_2O + \underline{1} PbCl_2\)
Balancing Pb: 1 on each side.
Balancing Cl: 2 on left, 2 on right.
Balancing H: 4 on left (2+2), 4 on right (22).
Balancing O: 2 on left, 2 on right.
Result: \(1 Pb(OH)_2 + 2 HCl \longrightarrow 2 H_2O + 1 PbCl_2\)

  1. Balancing Reactions 7 through 9

7) \(\underline{2} AlBr_3 + \underline{3} K_2SO_4 \longrightarrow \underline{6} KBr + \underline{1} Al_2(SO_4)_3\)
Balancing Al: 2 on right, 2 on left.
Balancing Br: 6 on left, 6 on right.
Balancing K: 6 on right, 6 on left (32).
Balancing SO4: 3 on left, 3 on right.
Result: \(2 AlBr_3 + 3 K_2SO_4 \longrightarrow 6 KBr + 1 Al_2(SO_4)_3\)

8) \(\underline{1} C_3H_8 + \underline{5} O_2 \longrightarrow \underline{3} CO_2 + \underline{4} H_2O\)
Balancing C: 3 on right, 3 on left.
Balancing H: 8 on right, 8 on left.
Balancing O: 10 on right (6+4), so 5 \(O_2\) on left.
Result: \(1 C_3H_8 + 5 O_2 \longrightarrow 3 CO_2 + 4 H_2O\)

9) \(\underline{1} FeCl_3 + \underline{3} NaOH \longrightarrow \underline{1} Fe(OH)_3 + \underline{3} NaCl\)
Balancing Fe: 1 on each side.
Balancing Cl: 3 on right, 3 on left.
Balancing Na: 3 on right, 3 on left.
Balancing OH: 3 on left, 3 on right.
Result: \(1 FeCl_3 + 3 NaOH \longrightarrow 1 Fe(OH)_3 + 3 NaCl\)

Examples

Chemical balancing is essential in industrial manufacturing, such as calculating the exact amounts of raw materials needed for large-scale production like synthesizing ammonia or producing polymers. By balancing equations, engineers ensure that reactants are used efficiently and that the resulting environmental output can be calculated precisely, preventing waste and ensuring safety. In everyday life, similar principles apply to baking—a recipe is essentially a balanced chemical "equation" where specific amounts of flour, sugar, and yeast must react in proportion to create the perfect loaf.

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