How much MnO2(s) should be added to excess HCl(aq) to obtain 215 mL of Cl2(g) at 25 °C and 705 Torr?
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To find out how much MnO2(s) should be added to excess HCl(aq) to obtain 215 mL of Cl2(g) at 25 °C and 705 Torr, you need to use stoichiometry and the ideal gas law.
Here's how you can solve it step by step:
1. Write the balanced chemical equation for the reaction between MnO2 and HCl:
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
2. Convert the given volume of Cl2 at 25 °C from mL to liters:
215 mL = 0.215 L
3. Use the ideal gas law equation to find the number of moles of Cl2:
PV = nRT
n = PV / RT
P = 705 Torr = 0.930 atm (1 atm = 760 Torr)
V = 0.215 L
R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
T = 25 °C = 298 K
n = (0.930 atm x 0.215 L) / (0.0821 L·atm/(mol·K) x 298 K)
4. Calculate the number of moles of Cl2. This represents the stoichiometric ratio between MnO2 and Cl2:
n(Cl2) = 4 x n(MnO2)
5. Convert the moles of Cl2 to moles of MnO2:
n(MnO2) = n(Cl2) / 4
6. Finally, calculate the mass of MnO2 using its molar mass:
Mass(MnO2) = n(MnO2) x Molar mass(MnO2)
By following these steps, you will be able to determine the amount of MnO2(s) needed.