Answer:
3.55 L.
Explanation:
We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:
Molar mass of MnO2 = 55 + (16×2)
= 55 + 32
= 87 g/mol
Mass of MnO2 = 12 g
Mole of MnO2 =...?
Mole = mass /Molar mass
Mole of MnO2 = 12 / 87
Mole of MnO2 = 0.138 mole
Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:
The balanced equation for the reaction is given below:
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
From the balanced equation above,
1 mole of MnO2 reacted to produce 1 mole of Cl2.
Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.
Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:
Temperature (T) = 25 °C = 25 °C + 273 = 298 K
Pressure (P) = 0.950 atm
Number of mole (n) = 0.138 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =.?
PV = nRT
0.950 × V = 0.138 × 0.0821 × 298
Divide both side by 0.950
V = (0.138 × 0.0821 × 298) / 0.950
V = 3.55 L
Therefore, 3.55 L of chlorine gas were obtained from reaction.
