Answer:
1.131g of MnO2
Explanation:
Step 1:
The balanced equation for the reaction.
MnO2(s) + 4HCl(aq)—> MnCl2(aq) + 2H2O(l) + Cl2(g)
Step 2:
Data obtained from the question. This includes:
Volume (V) of Cl2 = 295mL = 0.295L
Temperature (T) = 25°C = 25°C + 273 = 298K
Pressure (P) = 795 Torr = 795/760 = 1.05 atm
Number of mole (n) of Cl2 =.?
Gas constant (R) = 0.082atm.L/Kmol
Step 3:
Determination of the number of mole of Cl2 produced. The number of mole of Cl2 produced from the reaction can be obtained by applying the ideal gas equation as follow:
PV = nRT
1.05 x 0.295 = n x 0.082 x 298
Divide both side by 0.082 x 298
n = (1.05 x 0.295)/(0.082 x 298)
n = 0.013 mole.
Therefore, the number of mole of Cl2 produced is 0.013 mole.
Step 4:
Determination of number of mole of MnO2 needed to produce 0.013 mole of Cl2.
This is illustrated below:
From the balanced equation above,
1 mole of MnO2 produced 1 mole of Cl2.
Therefore, it will also take 0.013 mole of MnO2 to produce 0.013 mole of Cl2.
Step 5:
Converting 0.013 mole of MnO2 to grams. This is illustrated below:
Number of mole MnO2 = 0.013 mole
Molar Mass of MnO2 = 55 + (2x16) = 87g/mol
Mass of MnO2 =?
Mass = number of mole x molar Mass
Mass of MnO2 = 0.013 x 87
Mass of MnO2 = 1.131g
Therefore 1.131g of MnO2 should be added to excess HCl.
