[Azred]

Originally posted by Nanobyte: OK. So they want me to do molarity equations, yet they give no supporting evidence on HOW to do them.. Here they are:

1) What mass of each product results in 750. mL of 6.00 M H3PO4 reacts according to the following equation? 2H3PO4+3Ca(OH)2--->Ca3(PO4)2+6H20

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?

2) How many milliliters of 18.0 M H2SO4 are required to ract with 250. mL of 2.50 M Al(OH)3 if the products are aluminum sulfate and water?

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

3) 75.0 g of an AGNO3 solution reacts with enough Cu to produce 0.250 g of Ag by single replacement. What is the molarity of the initial AgNO3 solution if Cu(NO3)2 is the other product?

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.

I really only need the explanation for one, and then I should be able to figure the other two out. It's just that we have been sticking to specific molarity problems, and not placing them in with equations. Anyways, any help would be appreciated.

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