Answered

Practice Balancing Chemical Equations



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) $2 H_2 + \underline{1} O_2 \longrightarrow 2 H_2O$

6) $1 Pb(OH)_2 + \underline{2} HCl \longrightarrow 2 H_2O + 1 PbCl_2$

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



10) Solid sodium chloride and fluorine gas react to form solid sodium fluoride and chlorine gas.

AWrite a formula equation. (Include the physical state of each reactant and product.)

Verified Answer :

  • We verified that the provided equations (1-6) were already balanced according to the law of conservation of mass.
  • We balanced equations (7-9) by ensuring the number of atoms for each element was equal on both sides.
  • We balanced the final reaction between solid sodium chloride and fluorine gas, arriving at \(\boxed{2NaCl(s) + F_2(g) \longrightarrow 2NaF(s) + Cl_2(g)}\).

Explanation

  1. Problem Analysis and Balancing Strategy

To balance chemical equations, we must ensure that the number of atoms of each element is identical on both the reactant side (left) and the product side (right). This adheres to the law of conservation of mass. For equations 1-6, the equations were provided already balanced. For 7-9, we determine the coefficients by balancing atoms iteratively. For equation 10, we translate the words into a chemical equation and balance it.

  1. Verification of Equations 1-6

1) \(N_2 + 3H_2 \longrightarrow 2NH_3\): Balanced.
2) \(2Na + 2H_2O \longrightarrow 2NaOH + H_2\): Balanced.
3) \(2KClO_3 \longrightarrow 2KCl + 3O_2\): Balanced.
4) \(2NaCl + F_2 \longrightarrow 2NaF + Cl_2\): Balanced.
5) \(2H_2 + O_2 \longrightarrow 2H_2O\): Balanced.
6) \(Pb(OH)_2 + 2HCl \longrightarrow 2H_2O + PbCl_2\): Balanced.

  1. Balancing Equations 7-9

7) \(2AlBr_3 + 3K_2SO_4 \longrightarrow 6KBr + 1Al_2(SO_4)_3\). To balance this, we observe that 2 Aluminum atoms on the right require 2 \(AlBr_3\) on the left. This gives 6 Bromine atoms, requiring 6 \(KBr\) on the right. Finally, 3 Sulfur atoms on the right require 3 \(K_2SO_4\) on the left, which also balances the 6 Potassium atoms.
8) \(C_3H_8 + 5O_2 \longrightarrow 3CO_2 + 4H_2O\). We start with 3 Carbons (from \(C_3H_8\)) requiring 3 \(CO_2\), and 8 Hydrogens requiring 4 \(H_2O\). This results in 6 + 4 = 10 Oxygen atoms, needing 5 \(O_2\).
9) \(FeCl_3 + 3NaOH \longrightarrow Fe(OH)_3 + 3NaCl\). To balance 3 Chlorine atoms from \(FeCl_3\), we need 3 \(NaCl\). This forces the use of 3 \(NaOH\) on the left, which balances the 3 Hydroxide groups.

  1. Reaction of NaCl(s) and F2(g)

10) Solid sodium chloride and fluorine gas react to form solid sodium fluoride and chlorine gas.
Reactants: \(NaCl(s)\) and \(F_2(g)\).
Products: \(NaF(s)\) and \(Cl_2(g)\).
Unbalanced equation: \(NaCl(s) + F_2(g) \longrightarrow NaF(s) + Cl_2(g)\).
Balancing: We need 2 Sodium atoms on the left, requiring 2 \(NaCl\). This creates 2 Fluorine atoms, which is balanced by 1 \(F_2\). This also gives 2 Chlorine atoms, which is balanced by 1 \(Cl_2\).
Balanced equation: \(\boxed{2NaCl(s) + F_2(g) \longrightarrow 2NaF(s) + Cl_2(g)}\).

Examples

Chemical equations allow us to determine how much of each reactant is needed for a desired amount of product. For example, if you are industrial manufacturing chemicals like fertilizer or cleaning agents, you must calculate precise quantities to ensure safety and economic efficiency. By balancing equations, we ensure that resources are not wasted and harmful by-products are accounted for in a controlled reaction.

Related Questions