Question

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

_albr₃ + _k → _kbr + _al

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

_feo + _pdf₂ → _fef₂ + _pdo

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

_p₄ + _br₂ → ___pbr₃

Explanation:

Question 8

Step1: Balance Br atoms

On the left, \( \text{AlBr}_3 \) has 3 Br. On the right, \( \text{KBr} \) has 1 Br. So we need 3 \( \text{KBr} \) to balance Br. So coefficient of \( \text{KBr} \) is 3.

Step2: Balance K atoms

Since \( \text{KBr} \) has coefficient 3, K on the right is 3. So K on the left (from \( \text{K} \)) should be 3. Coefficient of \( \text{K} \) is 3.

Step3: Balance Al atoms

Al on left (from \( \text{AlBr}_3 \)) is 1, so Al on right (from \( \text{Al} \)) should be 1. Coefficient of \( \text{AlBr}_3 \) is 1, coefficient of \( \text{Al} \) is 1.

Step1: Check each atom

Fe: Left (from \( \text{FeO} \)) is 1, right (from \( \text{FeF}_2 \)) is 1.
O: Left (from \( \text{FeO} \)) is 1, right (from \( \text{PdO} \)) is 1.
Pd: Left (from \( \text{PdF}_2 \)) is 1, right (from \( \text{PdO} \)) is 1.
F: Left (from \( \text{PdF}_2 \)) is 2, right (from \( \text{FeF}_2 \)) is 2.
All atoms are balanced with coefficient 1 for each.

Step1: Balance P atoms

Left: \( \text{P}_4 \) has 4 P. Right: \( \text{PBr}_3 \) has 1 P. So we need 4 \( \text{PBr}_3 \) to balance P. Coefficient of \( \text{PBr}_3 \) is 4.

Step2: Balance Br atoms

\( \text{PBr}_3 \) has 4×3 = 12 Br. Left: \( \text{Br}_2 \) has 2 Br per molecule. So number of \( \text{Br}_2 \) is \( \frac{12}{2} = 6 \). Coefficient of \( \text{Br}_2 \) is 6.

Step3: Balance P atoms (check again)

\( \text{P}_4 \) has 4 P, \( \text{PBr}_3 \) has 4 P (since coefficient 4). So coefficient of \( \text{P}_4 \) is 1.

Answer:

1 \( \text{AlBr}_3 + \) 3 \( \text{K}
ightarrow \) 3 \( \text{KBr} + \) 1 \( \text{Al} \)

Question 9