[Nanobyte]

Originally posted by Azred: The 750 mL of 6.00 M H3PO4 is your limiting substance; assume that you have more than enough Ca(OH)2 to use up all the acid. As always, use volume * molarity to calculate the number of moles of the acid. Since mass cannot be destroyed, there must be just as much phosphate on both sides of the equation; from this, you can tell how much Ca3(PO4)2 you have. Did that make sense, without giving away the answer?

So, would you say: 6M H₃PO₄ x .75L solution= 4.5mol H₃PO₄ / 2mol H₃PO₄=2.25mol and then just multiply 2.25 to the products and figure the masses?

Calculate the number of moles of hydroxide (OH); you would need that many moles of hydrogen from the sulfuric acid to create water.

I do not understand how you can get the # of moles of one element from a compound. Would you get the mass of (OH)₃ multiply it to the mass you get from converting mL into grams? I'm still a little fuzzy on this one.

75 g of silver nitrate is .4414 moles (silver nitrate = 169.91); .250 g of Ag is .002317 moles and there would be just as many moles of nitrate. This would imply that there are .0011585 moles of Cu = .0736 g. hmmm....there seems to be one piece of information missing. Either the density of the silver nitrate solution or the volume of the copper nitrate solution after the reaction.

So, are you saying it can't be done?

[ 11-24-2002, 07:43 PM: Message edited by: Nanobyte ]