Note that the IUPAC approved name for that substance you call sodium trioxocarbonate(iv) is simply sodium carbonate.
Here is what is going on. The Zn reacts with the HCl to produce H2 and ZnCl2. All of the Zn reacts and some HCl is left unreacted. To determine how much of the HCl did not react it is treated with Na2CO3.
Zn + 2HCl ==> ZnCl2 + H2, then
2HCl + Na2CO3 ==> 2NaCl + H2O + CO2
mols HCl initially = M x L = 0.2 M x 0.100 L - 0.02 mols.
mols Na2CO3 used for 30 cc of the excess = M x L = 0.08 x 0.022 = 0.00176
Convert that to mols of HCl in the 30 cc sample taken. That will be 0.00176 x (2 mols HCl/1 mol Na2CO3) = 0.00176 x 2 = 0.00352.
Convert that to mols HCl in the 100 cc sample and that is
0.00352 x (100/30) = 0.0117.
How much HCl was used by the Zn? That's
0.02 mols HCl initially
-0.0117 mols HCl that did not react.
--------------------
0.00827 mols HCl used by the Zn.
mols Zn in the sample ix 0.00827 x (1 mol Zn/2 mol HCl) = 0.00827/2 = ?
Then grams Zn = mols Zn x molar mass Zn = ?
Post your work if you get stuck.
A piece of Zinc was added to 100cm³ of 0.2M of hydrochloric acid. After effervescene had stopped. 30cm³ of the resulting solution required 22cm³ of 0.08M of Sodium trioxocarbonate (iv) for complete reaction, calculate the mass of Zinc added
[Zn=65, H=1, C=12, Na=23, Cl=35.5]
1 answer
2 answers