12.5g of zinc trioxocarbonate (iv) znco3 were heated very strongly to a constant mass and the residue treated with excess hydrochloric acid HCl, calculate the mass of zinc chloride zncl2 that would be obtained. (zn=65 c=12 o =16 Cl=35.5)l
Please tell me who taught you to name ZnCO3 that name. It is NOT a correct name by any naming method. Did you learn this in class. Reading on your own. How. A correct name is zinc carbonate but there are others sanctioned by IUPAC.
ZnCO3 --> ZnO + CO2
mols ZnCO3 = g/molar mass = 12.5/113.4 = ?
Convert to mols ZnO.
?Moles ZnCO3 x (1 mol ZnO/1 mol ZnCO3) = xx, then
grams ZnO = mols ZnO x molar mass ZnO = yy.
Post your work if you get stuck.
Please let me know about the naming of ZnCO3.
To determine the mass of zinc chloride (ZnCl2) obtained, we first need to find the molar mass of zinc trioxocarbonate (IV) (ZnCO3):
ZnCO3 = (1 x Zn) + (1 x C) + (3 x O)
= (1 x 65) + (1 x 12) + (3 x 16)
= 65 + 12 + 48
= 125 g/mol
Next, we calculate the number of moles of ZnCO3 by dividing the given mass by the molar mass:
Number of moles = mass / molar mass
= 12.5 g / 125 g/mol
= 0.1 mol
From the balanced chemical equation, we know that 1 mole of ZnCO3 reacts with 2 moles of HCl to form 1 mole of ZnCl2:
ZnCO3 + 2HCl → ZnCl2 + H2O + CO2
Therefore, the number of moles of ZnCl2 produced would also be 0.1 mol.
Finally, we can calculate the mass of ZnCl2 formed using its molar mass:
Mass = number of moles x molar mass
= 0.1 mol x (1 x Zn + 2 x Cl)
= 0.1 mol x (1 x 65 + 2 x 35.5)
= 0.1 mol x (65 + 71)
= 0.1 mol x 136
= 13.6 g
Therefore, the mass of zinc chloride (ZnCl2) obtained would be 13.6 grams.
To calculate the mass of zinc chloride (ZnCl2) obtained from the reaction, we need to understand the stoichiometry of the reaction and use the given information.
First, let's write the balanced chemical equation for the reaction:
ZnCO3(s) + 2HCl(aq) -> ZnCl2(aq) + CO2(g) + H2O(l)
From this equation, we can see that one molecule of ZnCO3 reacts with two molecules of HCl to produce one molecule of ZnCl2.
Now, let's calculate the molar mass of ZnCO3:
Zn: 1 atom * 65 g/mol = 65 g/mol
C: 1 atom * 12 g/mol = 12 g/mol
3O: 3 atoms * 16 g/mol = 48 g/mol
Molar mass of ZnCO3 = 65 g/mol + 12 g/mol + 48 g/mol = 125 g/mol
We are given 12.5 g of ZnCO3, so we can calculate the number of moles of ZnCO3:
Number of moles = mass / molar mass = 12.5 g / 125 g/mol = 0.1 mol
Since the stoichiometry of the reaction shows that one mole of ZnCO3 produces one mole of ZnCl2, the number of moles of ZnCl2 produced will also be 0.1 mol.
Finally, let's calculate the mass of ZnCl2:
Molar mass of ZnCl2 = 65 g/mol + 2 * 35.5 g/mol = 136 g/mol
Mass of ZnCl2 = number of moles * molar mass = 0.1 mol * 136 g/mol = 13.6 g
Therefore, the mass of ZnCl2 obtained would be 13.6 grams.