Question

2.804 g of a mixture containing manganese (IV) oxide and concentrated hydrochloric acid was heated. The resultant mixture weighed 2.591g and 71cm cube of chlorine,measured at s.t.p., was given off. show that thi reaction obeys the law of conservation of mass.[1 mole gas at s.t.p. occupies 22.4dm cube].

Answer 1

Question

2.804 g of a mixture containing manganese (IV) oxide and concentrated hydrochloric acid was heated. The resultant mixture weighed 2.591 g and 71 cm³ of chlorine, measured at s.t.p., was given off. Show that the reaction obeys the law of conservation of mass. [1 mole gas at s.t.p. occupies 22.4 dm³].

Solution

The reaction equation is

$$\mathrm{MnO_2} + \mathrm{HCl} \rightarrow \mathrm{MnCl_2} + \mathrm{H_2O} + \mathrm{Cl_2}$$

Number of moles of Cl₂ (volume of Cl₂ should be converted to dm³):

$$n_{Cl_2} = \frac{V_{Cl_2}}{V_m} = \frac{0.071}{22.4} = 0.0032 \text{ moles}$$

Mass of Cl₂ (molar mass of Cl₂ = 70.9 g/mol):

$$m_{Cl_2} = n_{Cl_2} \cdot M_{Cl_2} = 0.0032 \cdot 70.9 = 0.213 \text{ g}$$

Mass of resultant mixture:

$$m_{rm} = 2.591 + 0.213 = 2.804 \text{ g}$$

As we can see that mass of resultant mixture is equal to the mass of initial mixture, so the reaction obeys the law of conservation of mass.