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.

Answer 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 $MnO_2:$  

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