Question

12.5g of zinc trioxocarbonate (iv) zoco3 where healed very strongly to a constant mass and the reduce treated with excess hydrochloric acid HCl. Calculate the mass of znco3 that could be obtained take Zn=65 c=12 o=16 h=1 cl=35.5

Answer 1

First let me point out that the International Union of Pure and Applied Chemistry (IUPAC) does not recognize the name you used for ZnCO3. They do recognize zinc carbonate. Second, let me retype the question to correct the typos/English/spelling.
"12.5g of zinc carbonate, <b>ZnCO3, WERE HEATED</B> very strongly to a constant mass and the <B> RESIDUE</B> treated with excess hydrochloric acid HCl. Calculate the mass of <b>ZnCO3</b> that could be obtained take Zn=65 c=12 o=16 h=1 cl=35.5"
ZnCO3 + heat ---> ZnO + CO2, then
ZnO + 2HCl ==> ZnCl2
moles ZnCO3 = grams/molar mass = 12.5 g/125.4 = 0.0997
Convert to mols ZnCl2 as follows:
0.0997 mols ZnCO3 X (1 mol ZnO/1 mol ZnCO3) x (1 mol ZnCl2/1 mol ZnO) = 0.0997 mols ZnCl2.
grams ZnCl2 x molar mass ZnCl2 = ? g