Answer to Question #22132 in Inorganic Chemistry for NBanner

Question

Hi,

the following includes my calculations of how to prepare a hydrochloric acid solution that will only dissolve 10% of a gram of calcium carbonate. Please take a look and tell me if I got it right and if not, tell we were did I went wrong. Thanks in advance!

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Number of moles in grams = mass (grams) / molar mass

Number of moles in 0,001 grams CaCO3 = 0.001 / 100 = 0.00001 mole

This will react with double the moles of HCl = 0.00001 x 2 = 0.00002 mole

Converting moles to grams of HCl needed:

HCl gram= molar mass x moles = 36.5 x 0.00002 = 0.00073 gram

0.00073 grams of HCl will react with 0,001 gram CaCO3

0.00073x1179 {density of HCl (22 Baume) in grams}=0.86 milliliter ~ 0.1 milliliter

0.1 milliliter of HCl will react with 0,001 gram CaCO3

0.01 milliliter of HCl will react with 0.0001 gram CaCO3 --------- the 10% per 1 mg

If all the previous calculations considered a 1N HCl solution

Dilution:

Concentration 1 x volume 1 = concentration two x volume 2

Answer 1

Number ofmoles in grams = mass (grams) / molar mass RIGHT
Number of moles in0,001 grams CaCO₃ = 0.001 / 100 = 0.00001 mole RIGHT
This will react withdouble the moles of HCl = 0.00001 x 2 = 0.00002 mole RIGHT
Converting moles tograms of HCl needed:
HCl gram= molar massx moles = 36.5 x 0.00002 = 0.00073 gram RIGHT
0.00073 grams of HClwill react with 0,001 gram CaCO₃ RIGHT
0.00073x1179 {densityof HCl (22 Baume) in grams}=0.86 milliliter ~ 0.1 milliliter RIGHT
0.1 milliliter of HClwill react with 0,001 gram CaCO₃ RIGHT
0.01 milliliter ofHCl will react with 0.0001 gram CaCO₃ --------- the 10% per 1 mg RIGHT
If all the previouscalculations considered a 1N HCl solution
Dilution:
Concentration 1 xvolume 1 = concentration two x volume 2 RIGHT