Question
A piece of zinc was added to 1000cm³ of 0.2m hydrochloric acid. After effervescence head stopped, 30cm³ of the resulting solution required 22cm³ of 0.08m sodium trioxocarbonate (iv) solution for complete neutralization, calculate the mass of zinc added.
Answers
If I wrote the formula from your name of sodium trioxocarbonate, I would write it as Na2O3CO3 and there is no such thing. Carbonate tells you that you want the [CO3]^2-. The trioxo is completely unnecessary. Also you wrote the concn of HCl as 0.2m. I'm sure you meant 0.2M. Also, I assume you meant 0.08 M Na2CO3.There is a difference between m and M. I will assume you meant to write 0.2 M.
Zn + 2HCl ==> ZnCl2 + H2.
mols HCl added initiall = M x L = 0.2 M x 1 L = 0.2 mol. How much was NOT used.
Na2CO3 + 2HCl ==> 2NaCl + CO2 + H2O
mols Na2CO3 not used = M x L = 0.08 x 0.022 L = 0.00176 mols. mols HCl = 2x that (from the coefficients) = 0.00352. That's for 30 cc of the solution. How much was not used of the original 1000 cc. That's 0.00352 x 1000/33 = 0.1173 mols.
How much was used in the Zn reaction. That's 0.2 mol - 0.1173 = 0.193 mols HCl used.
mols Zn used = half that from the coefficients then convert mols Zn to grams Zn. Post your work if you get stuck.
Zn + 2HCl ==> ZnCl2 + H2.
mols HCl added initiall = M x L = 0.2 M x 1 L = 0.2 mol. How much was NOT used.
Na2CO3 + 2HCl ==> 2NaCl + CO2 + H2O
mols Na2CO3 not used = M x L = 0.08 x 0.022 L = 0.00176 mols. mols HCl = 2x that (from the coefficients) = 0.00352. That's for 30 cc of the solution. How much was not used of the original 1000 cc. That's 0.00352 x 1000/33 = 0.1173 mols.
How much was used in the Zn reaction. That's 0.2 mol - 0.1173 = 0.193 mols HCl used.
mols Zn used = half that from the coefficients then convert mols Zn to grams Zn. Post your work if you get stuck.