Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s) should be added to excess HCl(aq) to obtain 205 mL Cl2(g) at 25 °C and 705 Torr ?

Answer 1

Answer:

0.676 grams of manganese (IV) oxide should be added.

Explanation:

Moles of chlorine gas = n

Volume of the chlorine gas = V = 205 mL = 0.205 L

Pressure of the chlorine gas = 705 Torr = $\frac{705}{760}atm=0.928 atm$

1 atm = 760 Torr

Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K

$PV=nRT$ ( ideal gs equation)

$n=\frac{PV}{RT}=\frac{0.928 atm\times 0.205 L}{0.0821 atm L/mol K\times 298 K}=0.00777 mol$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

According to reaction, 1 mole of chlorine gas is obtained from  1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :

$\frac{1}{1}\times 0.00777 mol=0.00777 mol$ of manganese (IV) oxide

Mass of 0.00777 moles of manganese (IV) oxide:

0.00777 mol × 87 g/mol = 0.676 g

0.676 grams of manganese (IV) oxide should be added.