HELP!!!! CHEM HOMEWORK! Consider the balanced chemical reaction shown below. 1 Ca3P2(s) + 6 H2O(l) 3 Ca(OH)2(s) + 2 PH3(g)?

Consider the balanced chemical reaction shown below.
Ca₃P₂(s) + 6H₂O(l) → 3Ca(OH)₂(s) + 2PH₃(g)

In a certain experiment, 6.629 g of Ca₃P₂(s) reacts with 4.323 g of H₂O(l) Ca₃P₂ is the limiting reactant.

How many grams of Ca(OH)₂(s) form?
__5.275__ g of Ca(OH)₂(s) form.

How many grams of PH₃(g) form?
__1.614__ g of PH₃(g) form.

How many grams of the excess reactant remains after the limiting reactant is completely consumed?
__4.063__ g of excess reactant remain

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Calculations :

Molar mass of Ca₃P₂ = (40.08 × 3 + 30.97×2) g/mol = 182.18 g/mol
Molar mass of H₂O = (1.01×2 + 16.00) g/mol = 18.02 g/mol
Molar mass of Ca(OH)₂ = (40.08 + 16.00×2 + 1.01×2) g/mol = 74.10 g/mol
Molar mass PH₃ = (30.97 + 1.01×3) g/mol = 34.00 g/mol

Ca₃P₂ is the limiting reactant.
No. of moles of Ca₃P₂ reacted = (4.323 g) / (182.18 g/mol) = 0.02373 mol

Mole ratio Ca₃P₂ : Ca(OH)₂ = 1 : 3
No. of moles of Ca(OH)₂ formed = (0.02373 mol) × 3 = 0.07119 mol
Mass of Ca(OH)₂ formed = (74.10 g/mol) × (0.07119 mol) = 5.275 g

Mole ratio Ca₃P₂ : PH₃ = 1 : 2
No. of moles of PH₃ formed = (0.02373 mol) × 2 = 0.04746 mol
Mass of PH₃ formed = (34.00 g/mol) × (0.04746 mol) = 1.614 g

H₂O is the excess reactant.
Mole ratio Ca₃P₂ : H₂O = 1 : 6
No. of moles of H₂O reacted = (0.02373 mol) × 6 = 0.1424 mol
Mass of H₂O reacted = (18.02 g/mol) × (0.1424 mol) = 2.566 g
Mass of H₂O remain = (6.629 - 2.566) g = 4.063 g

There is one major error in the answer which affects all the others values given.
When calculating the moles of Ca3P2 the person used the weight of water, namely, 4.323 g, rather
than the 6.629 g Ca3P2.