CaCO3 + 2HCl
What do you get for vol(HCl) needed? I keep getting 0.16dm^3 but from experimental evidence this is out by a factor of 10. It should be 0.016dm^3 I think.
What do you get for vol(HCl) needed? I keep getting 0.16dm^3 but from experimental evidence this is out by a factor of 10. It should be 0.016dm^3 I think.
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It is not possible to provide an answer for mL of HCl used because you have not provided the concentration of the HCl solution :
Molar mass CaCO3 = 100g/mol
4g CaCO3 = 4/100 = 0.04 mol CaCO3
This will require 0.04*2 = 0.08 mol HCl
If the concentration of the HCl solution is 5.0M , then you will require 0.08/5 = 0.016L OR more conveniently 16.0mL of the HCl solution .
But this is only correct if the HCl solution is 5.0M
You have not provided this essential information regarding the concentration of the HCl solution , so a totally confident answer cannot be given
Molar mass CaCO3 = 100g/mol
4g CaCO3 = 4/100 = 0.04 mol CaCO3
This will require 0.04*2 = 0.08 mol HCl
If the concentration of the HCl solution is 5.0M , then you will require 0.08/5 = 0.016L OR more conveniently 16.0mL of the HCl solution .
But this is only correct if the HCl solution is 5.0M
You have not provided this essential information regarding the concentration of the HCl solution , so a totally confident answer cannot be given
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mol(CaCO3) = mass/mr = 4/100 = 0.04 mol you need twice as many moles of HCl to neutalise so ...
mol (HCl) = 0.08 mol
now use mol = concentration * volume (to solve for volume)
This will give you a volume in dm^3 providing your concentration is in mol^-1dm^3 or (M)
mol (HCl) = 0.08 mol
now use mol = concentration * volume (to solve for volume)
This will give you a volume in dm^3 providing your concentration is in mol^-1dm^3 or (M)
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Why use dm3?, it's most common to use mm or cm3 = that could be your factor of 10