Equation balancing a. S(s) + O2(g) → SO3(8) b. 2Al(s) + 3C12(8) ► 2AICI,(s) E. 2NaOH(s) + H2SO,(ac) —Na, SO (ac) + H2O(1) d. CHg(8) + 50(8) - A 3C02(8) + 4H2O(g)

a. S(s) + 3/2 O₂(g) → SO₃(g)

b. Al(s) + 3/2 Cl₂(g) → AICI₃(s)

c. 2 NaOH(s) + H₂SO₄(ac) → Na₂SO₄(ac) + 2 H₂O(l)

d. C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) +  4 H₂O(g)

Explanation:

We will balance the equation using the trial and error method.

a. S(s) + O₂(g) → SO₃(g)

1. We will balance O atoms by multiplying O₂by 3/2.

S(s) + 3/2 O₂(g) → SO₃(g)

b. Al(s) + Cl₂(g) → AICI₃(s)

1. We will balance Cl atoms by multiplying Cl₂ by 3/2.

Al(s) + 3/2 Cl₂(g) → AICI₃(s)

c. NaOH(s) + H₂SO₄(ac) → Na₂SO₄(ac) + H₂O(l)

1. We will balance Na atoms by multiplying NaOH by 2.

2. We will balance H and O atoms by multiplying H₂O by 2.

2 NaOH(s) + H₂SO₄(ac) → Na₂SO₄(ac) + 2 H₂O(l)

d. C₃H₈(g) + O₂(g) → CO₂(g) +  H₂O(g)

1. We will balance C atoms by multiplying CO₂ by 3.

2. We will balance H atoms by multiplying H₂O by 4.

3. We will balance O atoms by multiplying O₂ by 5.

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) +  4 H₂O(g)

beta decay is the answer