Answer to Question #122896 in Inorganic Chemistry for Paul

Question

1.A piece of zinc was added to 1000cm3 of 0.2M hydrochloric acid. After effervescence had stopped, 28cm3 of the resulting solution required 17cm3 of 0.08M sodium trioxocarbonate IV solution for complete neutralization. Calculate the mass of the zinc added. (Zn = 65, HCl = 36.5, Na2CO3 = 106)

2.2g of a mixture of sodium hydroxide and sodium chloride (as impurity) were dissolved in 500cm3 of water. If 25cm3 of this solution were neutralized by 21cm3 of 0.1mol/dm3 hydrochloric acid, calculate the percentage of the sodium chloride impurity.

(NaOH = 40, HCl = 36.5, NaCl = 58.5)

Answer 1

The balanced chemical equations for the reactions are,

1.Zn(s) + 2HCl(aq)$\to$ ZnCl₂(aq) +H₂(g)

The other name of sodium trioxocarbonate (IV) is sodium carbonate.

2.Na₂CO₃(aq) +2HCl(aq) $\to$ 2NaCl(aq) +CO₂(g) +H₂O(l)

To get the concentration of excess HCl we use the second equation.

The reaction ratio is 1:2$\to$ 2:1:1

So, m(Na₂CO₃) reacts with m/2 (HCl)

$C1V1=\dfrac{C2}{2}V2$

$\dfrac{C2}{2}=\dfrac{C1V1}{V2}$

$=\dfrac{0.08M*0.017L}{0.028L}=0.0485$

Thus,

$C2=0.097M$

Calculate the number of moles of HCl that reacted with zinc

$m(HCl)=[C1HCl-C2HCl] *V1HCl$

$=[0.2-0.097]*1$

$m(HCl)=0.103mol$

Using equation 1 calculate the number of moles of zinc

1 mole of Zn react with 2 moles of HCl

$m(Zn)=\dfrac{m}{2}m(HCl)$

$Zn=\dfrac{1}{2}*0.103$

$=0.0515mol$

$Mass(Zn)=No.of moles(Zn)*Molar\space mass(Zn)$

$=0.0515*65\newline =3.3475$

$Zn\space(mass)\space is\space3.35g$