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: 2MnCO3 + O2=2MnO2 + 2CO2In the second step, manganese(IV) oxide and aluminum react to form manganese and aluminum oxidide: 3MnO2 + 4Al = 3Mn + 2Al2O3 Suppose the yield of the first step is 65.% and the yield of the second step is 80.%. Calculate the mass of manganese(II) carbonate required to make 8.0kg of manganese. Be sure your answer has a unit symbol, if needed, and is rounded to 2 significant digits.

Answer 1

Answer:

33 kg

Explanation:

Let's consider the two steps to obtain manganese.

Step 1: 2 MnCO₃ + O₂ = 2 MnO₂ + 2 CO₂

Step 2: 3 MnO₂ + 4 Al = 3 Mn + 2 Al₂O₃

The molar mass of Mn is 54.94 g/mol. The moles represented by 8.0 kg of Mn are:

8.0 × 10³ g × (1 mol / 54.94 g ) = 1.5 × 10² mol

In Step 2, the real yield of Mn is 1.5 × 10² mol and the percent yield is 80%. The theoretical yield is:

1.5 × 10² mol (Real) × (100 mol (Theoretical) / 80 mol (Real)) = 1.9 × 10² mol

According to Step 2, the molar ratio of Mn to MnO₂ is 3:3. The moles of MnO₂ are 3/3 × 1.9 × 10² mol = 1.9 × 10² mol

In Step 1, the real yield of MnO₂ is 1.9 × 10² mol and the percent yield is 65%. The theoretical yield is:

1.9 × 10² mol (Real) × (100 mol (Theoretical) / 65 mol (Real)) = 2.9 × 10² mol

According to Step 1, the molar ratio of MnO₂ to MnCO₃ is 2:2. The moles of MnCO₃ are 2/2 × 2.9 × 10² mol = 2.9 × 10² mol

The molar mass of MnCO₃ is 114.95 g/mol. The mass of MnCO₃ represented by 2.9 × 10² mol is:

2.9 × 10² mol × (114.95 g/mol) = 3.3 × 10⁴ g = 33 kg

Answer 2

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$.

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$.

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg