175g of Ca(OH)2 is 74.093 g/mol so...2.36 mols.
So if 2.36 mols of Ca(OH)2 is suppose to be used up completely, i need x amount of Sodium Carbonate now. Hmmm...I'll see what I can do with that
EDIT: Okay, Im kinda slow at this....Can you explain how I can solve for how many mols are needed to react with sodium carbonate now knowing that i have 2.36 mols of Calcium hydoxide?
EDIT 2: Nevermind, I figured it out on my own! Yay~ I feel smart now XD
Thanks for the help