Answer:
0.87g
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
MnO2(s) + 4HCl(aq) —> MnCl2(aq) + 2H2O(l) + Cl2(g)
Step 2:
Data obtained from the question. This includes the following:
Volume (V) of Cl2 obtained = 235mL
Temperature (T) = 25°C
Pressure (P) = 805 Torr
Step 3:
Conversion to appropriate unit.
For Volume:
1000mL = 1L
Therefore, 235mL = 235/1000 = 0.235L
For temperature:
Temperature (Kelvin) = temperature (celsius) + 273
Temperature (celsius) = 25°C
Temperature (Kelvin) = 25°C + 273 = 298K
For Pressure:
760 Torr = 1 atm
Therefore, 805 Torr = 805/760 = 1.06 atm
Step 4:
Determination of the number of mole of Cl2 produced. This is illustrated below:
The number of mole (n) of Cl2 produced can be obtained by using the ideal gas equation as follow:
PV = nRT
Volume (V) = 0.235L
Temperature (T) = 298k
Pressure (P) = 1.06 atm
Gas constant (R) = 0.082atm.L/Kmol
Number of mole (n)
PV = nRT
Divide both side by RT
n = PV /RT
n = (1.06 x 0.235)/(0.082 x 298)
n = 0.01 mole
Therefore 0.01 mole of Cl2 is produced from the reaction.
Step 5:
Determination of the number of mole MnO2 that produce 0.01 mole of Cl2. This is illustrated below:
MnO2(s) + 4HCl(aq) —> MnCl2(aq) + 2H2O(l) + Cl2(g)
From the balanced equation above,
1 mole of MnO2 produced 1 mole of Cl2.
Therefore, it will take 0.01 mole to MnO2 to also produce 0.01 mole of Cl2.
Step 6:
Converting 0.01 mole of MnO2 to grams.
This is illustrated below:
Number of mole MnO2 = 0.01 mole
Molar Mass of MnO2 = 55 + (2x16) = 87g/mol
Mass of MnO2 =?
Mass = number of mole x molar Mass
Mass of MnO2 = 0.01 x 87
Mass of MnO2 = 0.87g
Therefore, 0.87g of MnO2 is needed for the reaction.
