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Balance the following equations by entering the correct coefficient in each blank. Use 1 if the coefficient is 1.



10) $1 P_4 + \_\_ Br_2 \rightarrow \_\_ PBr_3$



11) $\_\_ LiCl + \_\_ Br_2 \rightarrow \_\_ LiBr + \_\_ Cl_2$



12) $\_\_ PbBr_2 + \_\_ HCl \rightarrow \_\_ HBr + \_\_ PbCl_2$

Verified Answer :

  • Equation 10: \(1 P_4 + 6 Br_2 \rightarrow 4 PBr_3\)
  • Equation 11: \(2 LiCl + 1 Br_2 \rightarrow 2 LiBr + 1 Cl_2\)
  • Equation 12: \(1 PbBr_2 + 2 HCl \rightarrow 2 HBr + 1 PbCl_2\)
  • \(\boxed{10) 6, 4; 11) 2, 1, 2, 1; 12) 1, 2, 2, 1}\)

Explanation

  1. Analysis of the Problem

The task is to balance three chemical equations. Balancing an equation requires that the number of atoms of each element on the reactant side must equal the number of atoms of each element on the product side. We apply this principle of the conservation of mass to each equation individually.

  1. Balancing Equation 10

For the equation \(1 P_4 + \underline{\quad} Br_2 \rightarrow \underline{\quad} PBr_3\):
We are given 1 mole of \(P_4\).

  1. Phosphorus atoms: There are 4 phosphorus atoms in \(P_4\), so we need 4 moles of \(PBr_3\) to balance the phosphorus. This gives the coefficient 4 for \(PBr_3\).
  2. Bromine atoms: Now we have \(4 \times 3 = 12\) bromine atoms on the product side. To have 12 bromine atoms on the reactant side, we need 6 moles of \(Br_2\) (since each \(Br_2\) molecule contains 2 bromine atoms).

Therefore, the balanced equation is \(1 P_4 + 6 Br_2 \rightarrow 4 PBr_3\).

  1. Balancing Equation 11

For the equation \(\underline{\quad} LiCl + \underline{\quad} Br_2 \rightarrow \underline{\quad} LiBr + \underline{\quad} Cl_2\):
Let the coefficients be \(a LiCl + b Br_2 \rightarrow c LiBr + d Cl_2\).

  1. Lithium (Li): \(a = c\)
  2. Chlorine (Cl): \(a = 2d\)
  3. Bromine (Br): \(2b = c\)

If we choose \(a = 2\), then \(c = 2\) (from lithium balance).
From the chlorine balance: \(2 = 2d \Rightarrow d = 1\).
From the bromine balance: \(2b = c = 2 \Rightarrow b = 1\).
The smallest integer coefficients are 2, 1, 2, 1.
The balanced equation is \(2 LiCl + 1 Br_2 \rightarrow 2 LiBr + 1 Cl_2\).

  1. Balancing Equation 12

For the equation \(\underline{\quad} PbBr_2 + \underline{\quad} HCl \rightarrow \underline{\quad} HBr + \underline{\quad} PbCl_2\):
Let the coefficients be \(a PbBr_2 + b HCl \rightarrow c HBr + d PbCl_2\).

  1. Lead (Pb): \(a = d\)
  2. Bromine (Br): \(2a = c\)
  3. Hydrogen (H): \(b = c\)
  4. Chlorine (Cl): \(b = 2d\)

If we choose \(a = 1\), then \(d = 1\) (from lead balance).
Then \(c = 2a = 2(1) = 2\) (from bromine balance).
Then \(b = c = 2\) (from hydrogen balance).
Check chlorine: \(b = 2d \Rightarrow 2 = 2(1)\), which is correct.
The balanced equation is \(1 PbBr_2 + 2 HCl \rightarrow 2 HBr + 1 PbCl_2\).

Examples

Chemical equations are used in industrial processes, like manufacturing fertilizers or refining metals, to ensure that the exact amounts of reactants are mixed to maximize production and minimize waste. By balancing equations, engineers can calculate the precise amounts of raw materials needed for a chemical reaction. This is essential for both efficiency and environmental safety. For example, knowing that \(1 PbBr_2 + 2 HCl \rightarrow 2 HBr + 1 PbCl_2\) tells a chemist that exactly 2 moles of hydrochloric acid are required to react with 1 mole of lead(II) bromide to produce the desired products.

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