More chemistry: molarity equations

[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

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?

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?

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.

[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.

It is never easy taking the first steps from theory to practice, but you'll get it. Worry not! [img]graemlins/awesomework.gif[/img]

[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 ]

[Azred]

Originally posted by Nanobyte: 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?

Yes. You would have 2.25 mol of the calcuim phosphate; use this to calculate the actual mass of calcium phosphate. Then, you can calculate the mass of only the calcium (mass of Ca/mass of calcium phosphate) to figure out how much calcium hydroxide is needed on the left, and finally the mass of the resulting water.
So far so good!

Originally posted by Nanobyte: 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.

.25 L * 2.5 M = .625 mole Al(OH₃), then multiply by 3 to get 1.875 mole OH. To make water, you need 1.875 mole H. You know the concentration of sulfuric acid and the number of moles; alter the formula to become volume (in L)= moles / concentration.

Originally posted by Nanobyte: So, are you saying it can't be done?

It does appear to be missing a piece of information. Go back to the original problem and make sure everything is there.
I did have a professor who would put problems on his test that were impossible to solve because you would be missing a piece of information; the correct answer to those problems was "not enough information given". What a sneak!

[Nanobyte]

Thanx Azred. Good thing is I only have three weeks left in this semester, and I wholeheartedly doubt that I'll have trouble with Earth&Atmosphere..

[Azred]

Don't mention it. I got my degree in chemistry back in 1991 and I haven't really ever used it. It's true--I really have forgotten more chemistry than most people ever know! *gasp* [img]graemlins/petard.gif[/img]