Chemistry Question Helllppppp?

Balanced equation for the reaction:
CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
Mole ratio CaH₂ : H₂O = 1 : 1

When CaH₂ completely reacts, H₂O needed = (0.18 mol) × 2 = 0.36 mol < 1.52 mol
Hence, H₂O is in excess, and CaH₂ is the limiting reactant which completely reacts.

According to the above equation, mole ratio CaH₂ : H₂ = 1 : 2
Moles of H₂ obtained = (0.18 mol) × 2 = 0.36 mol

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OR:

If CaH₂ is the limiting reactant:
(0.18 mol CaH₂ reacted) × (2 mol H₂ obtained / 1 mol CaH₂ reacted)
= 0.36 mol H₂ obtained

If H₂O is the limiting reactant:
(1.52 mol H₂O reacted) × (2 mol H₂ obtained / 2 mol H₂O reacted)
= 1.52 mol H₂ obtained

The limiting reactant produced smaller amounts of the products.
Hence, CaH₂ is the limiting reactant, and 0.36 mol H₂ is obtained.

CaH2 + 2 H2O → Ca(OH)2 + 2 H2

0,18 mole of CaH2 would react completely with 0,18 x (2/1) = 0,36 mole of H2O, but there is more H2O present than that, so H2O is in excess and CaH2 is the limiting reactant.

(0,18 mol CaH2) x (2 mol H2 / 1 mol CaH2) = 0,36 mol H2

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micatkie:  Mole ratio CaH₂ : H₂O ≠ 1 : 1