Question

1. balance the equation by typing the correct numbers in each blank. type in 1 if the coefficient is 1.

_licl + _br₂ → _libr + _cl₂

1. balance the equation by typing the correct numbers in each blank. type in 1 if the coefficient is 1.

_pbbr₂ + _hcl → _hbr + _pbcl₂

Explanation:

Question 11

Step1: Balance Cl atoms

On the left, we have 1 Cl in \( \text{LiCl} \), and on the right, we have 2 Cl in \( \text{Cl}_2 \). So we need to put a coefficient of 2 in front of \( \text{LiCl} \) to balance Cl: \( 2\text{LiCl} + \text{Br}_2
ightarrow \text{LiBr} + \text{Cl}_2 \)

Step2: Balance Li atoms

Now we have 2 Li on the left (from \( 2\text{LiCl} \)), so we need to put a coefficient of 2 in front of \( \text{LiBr} \) to balance Li: \( 2\text{LiCl} + \text{Br}_2
ightarrow 2\text{LiBr} + \text{Cl}_2 \)

Step3: Balance Br atoms

On the left, we have 2 Br in \( \text{Br}_2 \), and on the right, we have 2 Br in \( 2\text{LiBr} \). So Br is already balanced.

Step1: Balance Pb atoms

On the left, we have 1 Pb in \( \text{PbBr}_2 \), and on the right, we have 1 Pb in \( \text{PbCl}_2 \). So Pb is balanced for now.

Step2: Balance Cl atoms

On the left, we have 1 Cl in \( \text{HCl} \), and on the right, we have 2 Cl in \( \text{PbCl}_2 \). So we need to put a coefficient of 2 in front of \( \text{HCl} \) to balance Cl: \( \text{PbBr}_2 + 2\text{HCl}
ightarrow \text{HBr} + \text{PbCl}_2 \)

Step3: Balance Br atoms

On the left, we have 2 Br in \( \text{PbBr}_2 \), and on the right, we have 1 Br in \( \text{HBr} \). So we need to put a coefficient of 2 in front of \( \text{HBr} \) to balance Br: \( \text{PbBr}_2 + 2\text{HCl}
ightarrow 2\text{HBr} + \text{PbCl}_2 \)

Step4: Check H atoms

On the left, we have 2 H in \( 2\text{HCl} \), and on the right, we have 2 H in \( 2\text{HBr} \). So H is balanced.

Answer:

2, 1, 2, 1

Question 12