Chemistry Help?! Grams/Moles & Balancing Equations

This is the balanced equation :
2 H3PO4 + 3 Ca(OH)2 = Ca3(PO4)2 + 6 H20

Questions:
a. How many grams of Ca3(PO4)2 will be formed if 1.35 moles of H3PO4 are consumed in the above reaction?

b. How many moles of H20 will be formed when 0.518 moles of Ca(OH)2 are consumed in the above reaction?

c. How many grams of Ca(OH)2 are needed to react with 41.2 g of H3PO4 in the above reaction?

..... Unsure how to do this. HELP? :)

1H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2 + 6H2O
a) 1.35moles H3PO4 x (1Ca3(PO4)2 / 1H3PO4) = 1.35moles Ca3(PO4)2
1.35moles Ca3(PO4)2 x 309.94g/mole = 418.42g

b) 0.518moles Ca(OH)2 x (6H2O / 3Ca(OH)2) = 1.036moles

c) 41.2g H3PO4 / 97.97g/mole = 0.42moles
0.42moles H3PO4 x (3Ca(OH)2 / 2H3PO4) x 74g/mole Ca(OH)2 = 46.62g required

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