Is a review problem that i need help understanding step by step for tomorrows test

57g of Ca(OH)2 reacts with 46g of HCl. a) which one is the limiting reagent b) how many grams of the other reactant are left over? c) how many grams of CaCl2 are formed if the reaction goes to completion?

Ca(OH)2+ 2HCl-->CaCl2+2H2O

we have 57/ 74 mol of Ca(OH)2 and 46/ 36.5 mol of HCl

Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O
Mol: x 2x x 2x
comparing the amount of each we can see that a) HCl is the limiting reagent
Ca(OH)2 reacted 46/36.5 :2 = 46/73
b) Ca(OH)2 left over (57/74 - 46/73 ) * 74 = 10.369g
c) CaCl2 formed 46/73 * 111 = 69.945g

Answers will also depend on which stage of the calculation that you use rounding number for example 46/73 could be equal to 0.63 mol