Question

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(II) carbonate and oxygen react to form manganese(IV) oxide and carbon dioxide:
2MnCOA9 + 02(g) 2 MnO2(s) + 2 CO2(g)
In the second step, manganese(IV) oxide and aluminum react to form manganese and aluminum oxidide
3Mn02(6)+4Als)3Mnsl20s)
Suppose the yield of the first step is 66% and the yield of the second step is 97’%. Calculate the mass of manganese(II) carbonate required to make 4.0 kg of manganese.
Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Answers

  1. Answer:

    The answer is “6.52 kg and 13.1 kg“

    Explanation:

    For point a:  

    Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

    Equation:  

    3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

    Calculating the amount of MnCO_3

    = 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg

    For point b:

    Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

    =\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

    Equation:  

    3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

    Calculating the amount of MnO_2:  

    = 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\

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