Answer to Question #123028 in Chemistry for joseph

Question

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

Answer 1

Sodium trioxocarbonate (IV) is a name for sodium carbonate, Na₂CO₃.

C_total(HCl) = 0.2 M

V_total(HCl) = 1000 cm³ = 1 L

C(Na₂CO₃) = 0.08 M

V(Na₂CO₃) = 17 cm³ = 0.017 L

Solution:

Balanced chemical reactions:

(1): Zn(s) + 2HCl(aq) = ZnCl₂(aq) + H₂(g)

(2): Na₂CO₃(aq) + 2HCl(aq) = 2NaCl(aq) + CO₂(g) + H₂O(l)

Determine the concentration of excess HCl using equation (2):

According to the equation (2): n(Na₂CO₃) = n(HCl)/2

Thus:

2 × C(Na₂CO₃) × V(Na₂CO₃) = C_excess(HCl) × V_excess(HCl)

C_excess(HCl) = 2 × C(Na₂CO₃) × V(Na₂CO₃) / V_excess(HCl)

C_excess(HCl) = 2 × (0.08 M) × (0.017 L) / (0.028 L) = 0.097 M

Determine the amount of HCl that reacts with zinc:

n(HCl) = [C_total(HCl) - C_excess(HCl)] × V_total(HCl)

n(HCl) = (0.2 M - 0.097 M) × 1L = 0.103 mol

Determine the amount of Zn using equation (1):

According to the equation (1): n(Zn) = n(HCl)/2

Thus:

n(Zn) = (0.103 mol) / 2 = 0.0515 mol

Calculate the mass of Zinc:

m(Zn) = n(Zn) × M(Zn) = (0.0515 mol) × (65 g/mol) = 3.3475 g = 3.35 g

The mass of Zn is 3.35 g

Answer: The mass of Zn is 3.35 g