Question

1. apply concepts balance the following equations:

a. so₂(g) + o₂(g) —→ so₃(g)
b. fe₂o₃(s) + h₂(g) —→ fe(s) + h₂o(l)
c. p(s) + o₂(g) —→ p₄o₁₀(s)
d. al(s) + n₂(g) —→ aln(s)

Explanation:

Part a:

Step1: Count atoms on each side

• Left: S = 1, O = 2 (from \(SO_2\)) + 2 (from \(O_2\)) = 4
• Right: S = 1, O = 3 (from \(SO_3\))

Step2: Balance O first. Find least common multiple of 4 and 3, which is 12? Wait, better to adjust coefficients. Let's put 2 in front of \(SO_2\), 1 in \(O_2\), and 2 in \(SO_3\).

• New left: S = 2 (from \(2SO_2\)), O = 4 (2*2) + 2 (from \(O_2\)) = 6
• New right: S = 2 (from \(2SO_3\)), O = 6 (2*3)
• Now check S: left 2, right 2. O: left 6, right 6. Balanced.

Step1: Count atoms

• Left: Fe = 2, O = 3, H = 2 (from \(H_2\))
• Right: Fe = 1, O = 1 (from \(H_2O\)), H = 2 (from \(H_2O\))

Step2: Balance Fe first. Put 2 in front of Fe.

• Now right Fe = 2. Equation: \(Fe_2O_3(s) + H_2(g)

ightarrow 2Fe(s) + H_2O(l)\)

Step3: Balance O. Left O = 3, so put 3 in front of \(H_2O\).

• Now right O = 3. Equation: \(Fe_2O_3(s) + H_2(g)

ightarrow 2Fe(s) + 3H_2O(l)\)

Step4: Balance H. Right H = 6 (3*2), so put 3 in front of \(H_2\).

• Left H = 6 (3*2). Now check all: Fe=2, O=3, H=6. Balanced.

Step1: Count atoms

• Left: P = 1, O = 2 (from \(O_2\))
• Right: P = 4 (from \(P_4O_{10}\)), O = 10 (from \(P_4O_{10}\))

Step2: Balance P first. Put 4 in front of P.

• Now left P = 4. Equation: \(4P(s) + O_2(g)

ightarrow P_4O_{10}(s)\)

Step3: Balance O. Right O = 10, so put 5 in front of \(O_2\) (5*2=10).

• Now left O = 10 (5*2). Check all: P=4, O=10. Balanced.

Answer:

\(2SO_2(g) + O_2(g)
ightarrow 2SO_3(g)\)

Part b: