Asked by Maimunat
10.5g of zinc trioxocarbonate (iv) were heated very strongly to a constant mass and the residue treated with excess hydrochloric acid calculate the mass of zinc chloride that would be obtained
Zinc = 165, carbon = 12, oxygen =16, hydrogen=1 chlorine =35.5
Zinc = 165, carbon = 12, oxygen =16, hydrogen=1 chlorine =35.5
Answers
Answered by
DrBob222
First I want to ask where you were told that the name for ZnCO3 is zinc trioxocarbonate(iv). That is a made up name and it is not sanctioned by IUPAC. If you made it up let me know. If a teacher told you let me know. If you read it in some journal, book, newsletter, let me know. Also, note that Zn is not 165
ZnCO3 ==> ZnO + CO2
ZnO + 2HCl ==> ZnCl2 + H2O
mols ZnCO3 = grams/molar mass = about 10.5/125 = about 0.084
Now convert to mols ZnCl2 using the coefficients in the balanced equation.
0.084 mols ZnCO3 x (1 mol ZnO/1 mol ZnCO3) x (1 mol ZnCl2/1 mol ZnO) = 0.084 x 1/1 x 1/1 = 0.084 mols ZnCl2
Then grams ZnCl2 = mols ZnCl2 x molar mass ZnCl2.
Post your work if you don't understand anything about this.
ZnCO3 ==> ZnO + CO2
ZnO + 2HCl ==> ZnCl2 + H2O
mols ZnCO3 = grams/molar mass = about 10.5/125 = about 0.084
Now convert to mols ZnCl2 using the coefficients in the balanced equation.
0.084 mols ZnCO3 x (1 mol ZnO/1 mol ZnCO3) x (1 mol ZnCl2/1 mol ZnO) = 0.084 x 1/1 x 1/1 = 0.084 mols ZnCl2
Then grams ZnCl2 = mols ZnCl2 x molar mass ZnCl2.
Post your work if you don't understand anything about this.
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