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

The answer is "6.52 kg and 13.1 kg"

Step-by-step explanation:

For point a:

`Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = (Actual \ yield)/(Theoretical\ yield) * 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = (Actual \ yield)/(Percent \ yield) * 100\%\\\\=(6.52 \ kg)/(66 \%) * 100\% =9.88 \ kg\\\\`

Equation:

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

Calculating the amount of
`MnCO_3`

`= 9.88 \ kg * (1000 \ g)/(1 \ kg) * (1 \ mol \ MnO_2)/(86.94 \ g) * (2 \ Mol \ MnCO_3)/(2 \ mol \ MnO_2) * (114.95\ g)/(1 \ mol \ MnCO_3 )* (1\ kg)/(1000\ g)\\\\= 13.1 \ kg`

For point b:

`Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = (Actual \ yield)/(Theoretical \ yield) * 100 \% \\\\Theoretical \ yield\ of\ Mn = (Actual \ yield)/(Percent \ yield) * 100\%\\\\`

`=(4.0 \ kg)/(97.0\%) * 100\% =4.12 \ kg`

Equation:

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

Calculating the amount of
`MnO_2:`

`= 4.12 \ kg * (1000 \ g)/(1 \ kg) * (1 \ mol \ Mn)/(54.94 \ g) * (3 \ Mol \ MnO_2)/(3 \ mol \ Mn) * (86.94 \ g)/(1 \ mol \ MnO_2 )\\\\= 6516 \ g \\\\=6.52 \ kg\\\\`