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C asked in Science & MathematicsChemistry · 5 years ago

Chemistry Question... I am stuck on this... I don t know if I'm on the right track. PLEASE HELP? My teacher doesn't lecture & I'm confused?

KMnO4 reacts with hydrochloric acid to give us chlorine gas.

a) If 8.5 g of KMnO4 is mixed with 50 mL of 6.0 M HCl, what is the final pressure of chlorine gas that could be generated if the final temperature is 110 degrees celsius in a 1.5 L flask?

I know the balanced equation is:

2 KMnO4 + 16 HCl -----> 2 MnCl2 + 2 KCl + 5Cl2 (g) + 8H2O

I also think that due to STP the temperature should be in Kelvin for this type of problem, but I am still stuck and my book wasn't helping.

b) How much of the other reagent (the one that is not the limiting reactant) is left over?

THANKS SO MUCH!!!

1 Answer

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  • Trevor H
    Lv 7
    5 years ago
    Favorite Answer

    Molar mass of KMnO4 is 158.03 g/mol

    mol KMnO4 in 8.5g = 8.5/158.03 = 0.0538mol KMnO4

    Mol HCl in 50mL of 6.0M HCl solution = 50/1000*6.0 = 0.30 mol HCl

    From the equation:

    2mol KMnO4 will react with 16 mol HCl

    0.0538 mol KMnO4 will react with 0.0538*16/2 = 0.4304 mol HCl

    You do not have enough HCl to react with all the KMnO4 . The HCl is limiting

    From the equation:

    16 mol HCl will produce 5mol Cl2

    Mol Cl2 produced by 0.30 mol HCl = 0.3*5/16 = 0.0937 mol Cl2 produced.

    Use gas equation to calculate pressure in flask

    PV = nRT

    P = ???

    V = 1.5L

    n = 0.0937

    R = 0.082057

    T = 110+273 = 383K

    P*1.5 = 0.0937*0.082057*383

    P = 0.0937*0.082057*383 / 1.5

    P = 1.96 atm - that is the pressure.

    b) How much KMnO4 remains:

    Calculated above : the 0.0538 mol KMnO4 reacts fully with 0.4304 mol HCl

    But only 0.3 mol HCl reacted

    Mol KMnO4 reacted = 0.0538* 0.3/0.4304 = 0.0375 mol KMnO4 reacted

    Mol KMnO4 unreacted = 0.0538 - 0.0375 = 0.0163 mol KMnO4 unreacted

    Mass of KMnO4 remaining unreacted = 0.0163 * 158.03 = 2.5g KMnO4 left over

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