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Balance each equation by typing the correct numbers in each blank. Type in 1 if the coefficient is 1.



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



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



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

Verified Answer :

  • The first equation \(1 P_4 + x Br_2 \rightarrow y PBr_3\) balances as \(1 P_4 + 6 Br_2 \rightarrow 4 PBr_3\).
  • The second equation \(a LiCl + b Br_2 \rightarrow c LiBr + d Cl_2\) balances as \(2 LiCl + 1 Br_2 \rightarrow 2 LiBr + 1 Cl_2\).
  • The third equation \(a PbBr_2 + b HCl \rightarrow c HBr + d PbCl_2\) balances as \(1 PbBr_2 + 2 HCl \rightarrow 2 HBr + 1 PbCl_2\).
  • Final answers: \(\boxed{6, 4}\), \(\boxed{2, 1, 2, 1}\), \(\boxed{1, 2, 2, 1}\).

Explanation

  1. Balancing the First Equation

To balance a chemical equation, we must ensure that the number of atoms of each element is identical on both the reactant and product sides, adhering to the law of conservation of mass. We analyze each equation one by one.

For the first equation, we have \(1 P_4 + x Br_2 \rightarrow y PBr_3\).

  1. Phosphorous (P) atoms: There are 4 on the left, so we need 4 on the right: \(y = 4\).
  2. Bromine (Br) atoms: With \(y = 4\), we have \(4 \times 3 = 12\) atoms on the right. To balance this on the left, \(2x = 12\), which means \(x = 6\).

The balanced equation is \(1 P_4 + 6 Br_2 \rightarrow 4 PBr_3\).

  1. Balancing the Second Equation

For the second equation, we have \(a LiCl + b Br_2 \rightarrow c LiBr + d Cl_2\).

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

Letting \(d = 1\), we find \(a = 2\), and since \(a = c\), \(c = 2\). From the bromine balance, \(2b = c = 2\), so \(b = 1\).
The balanced equation is \(2 LiCl + 1 Br_2 \rightarrow 2 LiBr + 1 Cl_2\).

  1. Balancing the Third Equation

For the third equation, we have \(a PbBr_2 + b HCl \rightarrow c HBr + d PbCl_2\).

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

Letting \(d = 1\), we get \(a = 1\). Then \(c = 2a = 2\), and \(b = c = 2\). Finally, checking chlorine: \(b = 2d \implies 2 = 2(1)\), which is correct.
The balanced equation is \(1 PbBr_2 + 2 HCl \rightarrow 2 HBr + 1 PbCl_2\).

Examples

Balancing chemical equations is essential in chemistry for determining the precise amounts of reactants needed and products produced in a reaction. For example, if you are working in a lab to produce a specific compound, knowing the exact stoichiometric ratios ensures you do not waste reagents. If a process requires 1 mole of \(PbBr_2\), you know you need exactly 2 moles of \(HCl\) to react completely, producing 2 moles of \(HBr\) and 1 mole of \(PbCl_2\). This skill allows for efficiency and safety in both industrial manufacturing and research experiments.

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