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Balance the following chemical equations:



1) $1 N_2 + 3 H_2 \longrightarrow 2 NH_3$

2) $2 Na + 2 H_2 O \longrightarrow 2 NaOH + 1 H_2$

3) $2 KClO_3 \longrightarrow 2 KCl + 3 O_2$

4) $2 NaCl + 1 F_2 \longrightarrow 2 NaF + 1 Cl_2$

5) $2 H_2 + 1 O_2 \longrightarrow 2 H_2 O$

6) $\_ Pb(OH)_2 + \_ HCl \longrightarrow \_ H_2 O + \_ PbCl_2$

7) $\_ AlBr_3 + \_ K_2 SO_4 \longrightarrow \_ KBr + \_ Al_2(SO_4)_3$

8) $\_ C_3 H_8 + \_ O_2 \longrightarrow \_ CO_2 + \_ H_2 O$

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

Verified Answer :

  • To balance the equations, we apply the law of conservation of mass by ensuring the number of atoms for each element is equal on both sides.
  • Equation 6 balances as \(1 Pb(OH)_2 + 2 HCl \longrightarrow 2 H_2O + 1 PbCl_2\).
  • Equation 7 balances as \(2 AlBr_3 + 3 K_2SO_4 \longrightarrow 6 KBr + 1 Al_2(SO_4)_3\).
  • Equation 8 balances as \(1 C_3H_8 + 5 O_2 \longrightarrow 3 CO_2 + 4 H_2O\).
  • Equation 9 balances as \(1 FeCl_3 + 3 NaOH \longrightarrow 1 Fe(OH)_3 + 3 NaCl\).

Explanation

  1. Problem Analysis and Objectives

The task is to balance four chemical equations by finding the appropriate stoichiometric coefficients. The law of conservation of mass states that the number of atoms of each element must remain the same on both the reactant and product sides. We will determine these coefficients for equations 6, 7, 8, and 9 by setting up systems of equations based on atom counts.

  1. Balancing Equation 6

Equation 6 is \(1 Pb(OH)_2 + x HCl \longrightarrow y H_2O + z PbCl_2\).

  • Lead (Pb): \(1 = z\), so \(z = 1\).
  • Chlorine (Cl): \(x = 2z = 2(1) = 2\), so \(x = 2\).
  • Hydrogen (H): \(2 + x = 2y\). Substituting \(x=2\): \(2 + 2 = 2y\), so \(4 = 2y\), which gives \(y = 2\).
  • Oxygen (O): \(2 = y\), which is consistent with \(y = 2\).

The balanced equation is: \(1 Pb(OH)_2 + 2 HCl \longrightarrow 2 H_2O + 1 PbCl_2\).

  1. Balancing Equation 7

Equation 7 is \(x AlBr_3 + y K_2SO_4 \longrightarrow z KBr + w Al_2(SO_4)_3\).

  • Aluminum (Al): \(x = 2w\).
  • Bromine (Br): \(3x = z\).
  • Potassium (K): \(2y = z\).
  • Sulfur (S): \(y = 3w\).

Let \(w = 1\). Then \(x = 2(1) = 2\). Since \(x = 2\), \(z = 3x = 6\). Since \(z = 6\), \(y = z/2 = 3\).
Checking Oxygen (O): \(4y = 4(3) = 12\) and \(12w = 12(1) = 12\). It matches.
The balanced equation is: \(2 AlBr_3 + 3 K_2SO_4 \longrightarrow 6 KBr + 1 Al_2(SO_4)_3\).

  1. Balancing Equation 8

Equation 8 is \(x C_3H_8 + y O_2 \longrightarrow z CO_2 + w H_2O\).

  • Carbon (C): \(3x = z\).
  • Hydrogen (H): \(8x = 2w\), so \(w = 4x\).
  • Oxygen (O): \(2y = 2z + w\). Substituting \(z=3x\) and \(w=4x\): \(2y = 2(3x) + 4x = 6x + 4x = 10x\), so \(y = 5x\).

Let \(x = 1\). Then \(z = 3, w = 4, y = 5\).
The balanced equation is: \(1 C_3H_8 + 5 O_2 \longrightarrow 3 CO_2 + 4 H_2O\).

  1. Balancing Equation 9

Equation 9 is \(x FeCl_3 + y NaOH \longrightarrow z Fe(OH)_3 + w NaCl\).

  • Iron (Fe): \(x = z\).
  • Chlorine (Cl): \(3x = w\).
  • Sodium (Na): \(y = w\), so \(y = 3x\).
  • Oxygen (O): \(y = 3z\). Substituting \(x=z\): \(y = 3z = 3x\), which matches the Na count.

Let \(x = 1\). Then \(z = 1, y = 3, w = 3\).
The balanced equation is: \(1 FeCl_3 + 3 NaOH \longrightarrow 1 Fe(OH)_3 + 3 NaCl\).

  1. Conclusion and Final Results

The balanced equations with their respective stoichiometric coefficients are:
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) \(1 FeCl_3 + 3 NaOH \longrightarrow 1 Fe(OH)_3 + 3 NaCl\)
All equations are now balanced according to the law of conservation of mass.

Examples

Balancing chemical equations is fundamental in industrial processes like pharmaceutical synthesis or material manufacturing, where precisely measuring reactants is necessary for the desired yield. For example, to manufacture a specific amount of medication, a chemist needs to know exactly how much reagent to add based on the balanced stoichiometric ratios so that no excess waste is created and the product is pure. Understanding these ratios allows scientists to scale up laboratory-scale experiments to industrial-level production efficiently.

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