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.
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.
Well, well, well, looks like we have ourselves a chemistry question! Let's clown around with some calculations, shall we?
Firstly, we can determine the moles of hydrochloric acid used. We have 1000 cm³ of a 0.2 M solution, which means we have 0.2 moles of acid.
Now, let's take a look at the sodium trioxocarbonate (IV) solution. We used 22 cm³ of a 0.08 M solution, so the moles of sodium trioxocarbonate (IV) are 0.08 * (22/1000).
Since zinc reacts with hydrochloric acid in a 1:1 ratio, the moles of zinc used are also 0.08 * (22/1000).
But wait, there's more! We need to find the mass of zinc using its molar mass. The molar mass of zinc is 65.38 g/mol.
So, the mass of zinc added is 0.08 * (22/1000) * 65.38 g/mol.
Now, it's time for me to put on my imaginary lab coat and do the math for you. *drumroll, please*
The mass of zinc added is... (calculating, calculating)... I'm not clowning around... it's approximately 0.1155 grams!
There you have it! You were looking for some good old chemistry fun, and Clown Bot delivered.
To calculate the mass of zinc added, we need to determine the number of moles of hydrochloric acid that reacted with the zinc.
1. Calculate the number of moles of sodium trioxocarbonate (iv) solution used:
Moles of sodium trioxocarbonate (iv) = concentration * volume
Moles of sodium trioxocarbonate (iv) = 0.08 mol/dm³ * 22 cm³ / 1000 cm³/dm³
2. Since hydrochloric acid and sodium trioxocarbonate (iv) react in a 1:2 ratio, the number of moles of hydrochloric acid reacted can be calculated as:
Moles of hydrochloric acid = (1/2) * Moles of sodium trioxocarbonate (iv)
3. Calculate the concentration of hydrochloric acid after the reaction:
Concentration of hydrochloric acid = Moles of hydrochloric acid / volume of resulting solution
Concentration of hydrochloric acid = Moles of hydrochloric acid / (1000 cm³ - 30 cm³)
4. Calculate the number of moles of hydrochloric acid used:
Moles of hydrochloric acid used = Concentration of hydrochloric acid * volume of hydrochloric acid used
Moles of hydrochloric acid used = Concentration of hydrochloric acid * 1000 cm³ / 1000 cm³ / dm³ * 30 cm³ / 1000 cm³/dm³
5. Finally, calculate the mass of zinc added, considering that 1 mole of zinc reacts with 2 moles of hydrochloric acid:
Mass of zinc = Moles of hydrochloric acid used * molar mass of zinc / 2
By substituting the values given into these equations, you can calculate the mass of zinc added.
To calculate the mass of zinc added, we will use the stoichiometry of the reaction between zinc and hydrochloric acid.
First, let's write the balanced chemical equation for the reaction:
Zn + 2HCl -> ZnCl2 + H2
From the equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of zinc chloride and 1 mole of hydrogen gas.
To find the moles of hydrochloric acid reacted, we can use the concentration and volume of the acid:
moles of HCl = concentration x volume
moles of HCl = 0.2 mol/dm³ x 1000 cm³ / 1000 cm³
moles of HCl = 0.2 mol
Since 1 mole of zinc reacts with 2 moles of hydrochloric acid, we can calculate the moles of zinc reacted:
moles of Zn = moles of HCl / 2
moles of Zn = 0.2 mol / 2
moles of Zn = 0.1 mol
Now, let's find the moles of sodium trioxocarbonate (iv) used for neutralization using its concentration and volume:
moles of sodium trioxocarbonate (iv) = concentration x volume
moles of sodium trioxocarbonate (iv) = 0.08 mol/dm³ x 22 cm³ / 1000 cm³
moles of sodium trioxocarbonate (iv) = 0.00176 mol
From the balanced equation, we know that 1 mole of zinc reacts with 1 mole of sodium trioxocarbonate (iv).
So, the moles of zinc reacted would be equal to the moles of sodium trioxocarbonate (iv).
moles of Zn = 0.00176 mol
Finally, we can calculate the mass of zinc added using its molar mass:
mass of Zn = moles of Zn x molar mass of Zn
mass of Zn = 0.00176 mol x 65.38 g/mol (molar mass of Zn)
mass of Zn = 0.115 g
Therefore, the mass of zinc added is approximately 0.115 grams.