Question
A piece of zinc is added to 1000cm3 of 0.2mol/dm3 hydrochloric acid. After effervescence had stopped, 28cm3 of the resulting solution required 17cm3 of 0.08mol/dm3 sodium trioxocarbonate(IV) solution for complete neutralization.
Calculate the mass of the zinc added. [Zn =65; H=1; Cl=35.5; Na=23; C=12; O=16]
Calculate the mass of the zinc added. [Zn =65; H=1; Cl=35.5; Na=23; C=12; O=16]
Answers
2HCl + Zn ==> ZnCl2 + H2
0.2 mol/dm3 = 0.2 mol/L = 0.2 M and 1000 cm3 = 1 L
How many mols HCl were there initially? That's M x L = 0.2 x 1 = 0.2 mol.
Not all of the HCl reacted. How much was left?
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 = M x L = 0.08 x 0.017 = 0.00136
mols HCl must be 2*0.00136 = 0.00272 in the 28 cc aliquot.
How much was left in the 1000 cc? That's 0.00272 x (1000/28) = 0.0971 mols HCl that were not used. How many mols were used? That's
0.2 mol initially - 0.0971 = 0.1029
Referring to the equation with Zn + 2HCl, mols Zn must be 1/2 x 0.0971 = ?
Grams Zn will be grams = mols x atomic mass Zn = ?
Post your work if you get stuck.
0.2 mol/dm3 = 0.2 mol/L = 0.2 M and 1000 cm3 = 1 L
How many mols HCl were there initially? That's M x L = 0.2 x 1 = 0.2 mol.
Not all of the HCl reacted. How much was left?
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 = M x L = 0.08 x 0.017 = 0.00136
mols HCl must be 2*0.00136 = 0.00272 in the 28 cc aliquot.
How much was left in the 1000 cc? That's 0.00272 x (1000/28) = 0.0971 mols HCl that were not used. How many mols were used? That's
0.2 mol initially - 0.0971 = 0.1029
Referring to the equation with Zn + 2HCl, mols Zn must be 1/2 x 0.0971 = ?
Grams Zn will be grams = mols x atomic mass Zn = ?
Post your work if you get stuck.