Balanced cracking equation with two distinct alkenes and another product

Steps:

• Confirm decane formula C10H22 and cracking produces alkenes (CnH2n) plus possibly H2.
• Balance atoms in each equation: carbons must total 10, hydrogens 22.
• Identify alkenes: count distinct types (different carbon chains).
• Check for exactly two different alkenes and at least one non-alkene product.

Why D is correct:

• Balances C (10=2+8) and H (22=2+4+16) per conservation of mass; yields two alkenes (ethene C2H4, octene C8H16) and H2.

Why the others are wrong:

• A: Hydrogens unbalanced (20 vs 22); three alkenes, no other product.
• B: Three different alkenes (C2H4, C3H6, C5H10), exceeds two.
• C: Carbons unbalanced (8 vs 10); only two alkenes but invalid equation.

Final answer: D