I think that I have about 50 mL of HCl and I need to know how much CaCo3 is needed to neutralize the HCl perferably in grams or moles. Thanks!
2007-04-03 16:40:01 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Opps it actually is 6 M HCl...Sorry!
2007-04-03 17:45:24 · update #1
It depends on the concentration of the HCl. If you're using concentrated, which is 12 M, then your 50 ml of HCl would have 0.6 moles, which would need 0.3 moles of CaCO3 to neutralize, which is 30 g of CaCO3 (MW CaCO3 = 100).
2007-04-03 16:44:14 · answer #1 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋
Well, 1st I would need to know the molarity of the HCl solution. Basically, the balanced equation is:
CaCO3 + 2 HCl --> CaCl2 + H2CO3, so for every 2 moles of HCl, you would need 1 mole of CaCO3.
2007-04-03 16:47:01 · answer #2 · answered by misoma5 7 · 0⤊ 0⤋
You can't do this without knowing the concentration or the molarity of the HCl.
2007-04-03 16:44:25 · answer #3 · answered by CornellAdamO 3 · 0⤊ 0⤋
50ml= .050L
mol=L*M = .05L*6.M =.3mol HCl
.3mol HCl(1molCaCO3/2molHCl)
(100gCaCO3/1molCaCO3)=
15gCaCO3
Good luck
2007-04-03 20:37:38 · answer #4 · answered by bige1236 4 · 0⤊ 0⤋