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]

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.

To find the mass of zinc added, we need to calculate the number of moles of zinc involved in the reaction.

First, let's determine the number of moles of hydrochloric acid in the solution. The concentration of the hydrochloric acid is given as 0.2 mol/dm3 (moles per liter) and the volume is 1000 cm3. However, we need to convert the volume from cm3 to dm3, so we divide the volume by 1000:

Volume of hydrochloric acid = 1000 cm3 ÷ 1000 = 1 dm3

Now we can calculate the number of moles of hydrochloric acid using the formula:

Moles = Concentration × Volume

Moles of hydrochloric acid = 0.2 mol/dm3 × 1 dm3 = 0.2 moles

From the chemical equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to form zinc chloride and hydrogen gas. Therefore, the number of moles of zinc involved in the reaction is half the number of moles of hydrochloric acid.

Moles of zinc = 0.2 moles ÷ 2 = 0.1 moles

Next, we need to calculate the number of moles of sodium trioxocarbonate(IV) solution used to neutralize the resulting solution. The volume of the sodium trioxocarbonate(IV) solution used is given as 17 cm3, but we need to convert this to dm3:

Volume of sodium trioxocarbonate(IV) solution = 17 cm3 ÷ 1000 = 0.017 dm3

Now, we can calculate the number of moles of sodium trioxocarbonate(IV) solution using the formula:

Moles = Concentration × Volume

Moles of sodium trioxocarbonate(IV) solution = 0.08 mol/dm3 × 0.017 dm3 = 0.00136 moles

According to the chemical equation, 1 mole of sodium trioxocarbonate(IV) solution reacts with 1 mole of zinc to form zinc carbonate and sodium chloride. Therefore, the number of moles of zinc used is equal to the number of moles of sodium trioxocarbonate(IV) solution used.

Moles of zinc = 0.00136 moles

Finally, we can calculate the mass of the zinc added using its molar mass, which is given as 65 g/mol:

Mass of zinc = Moles × Molar mass

Mass of zinc = 0.00136 moles × 65 g/mol = 0.0884 g

Therefore, the mass of the zinc added is approximately 0.0884 grams.