Respuesta :
Answer : The mass of [tex]MnCO_3[/tex] required are, 35 kg
Explanation :
First we have to calculate the mass of [tex]MnO_2[/tex].
The first step balanced chemical reaction is:
[tex]2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2[/tex]
Molar mass of [tex]MnCO_3[/tex] = 115 g/mole
Molar mass of [tex]MnO_2[/tex] = 87 g/mole
Let the mass of [tex]MnCO_3[/tex] be, 'x' grams.
From the balanced reaction, we conclude that
As, [tex](2\times 115)g[/tex] of [tex]MnCO_3[/tex] react to give [tex](2\times 87)g[/tex] of [tex]MnO_2[/tex]
So, [tex]xg[/tex] of [tex]MnCO_3[/tex] react to give [tex]\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg[/tex] of [tex]MnO_2[/tex]
And as we are given that the yield produced from the first step is, 65 % that means,
[tex]60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg[/tex]
The mass of [tex]MnO_2[/tex] obtained = 0.4542x g
Now we have to calculate the mass of [tex]Mn[/tex].
The second step balanced chemical reaction is:
[tex]3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3[/tex]
Molar mass of [tex]MnO_2[/tex] = 87 g/mole
Molar mass of [tex]Mn[/tex] = 55 g/mole
From the balanced reaction, we conclude that
As, [tex](3\times 87)g[/tex] of [tex]MnO_2[/tex] react to give [tex](3\times 55)g[/tex] of [tex]Mn[/tex]
So, [tex]0.4542xg[/tex] of [tex]MnO_2[/tex] react to give [tex]\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg[/tex] of [tex]Mn[/tex]
And as we are given that the yield produced from the second step is, 80 % that means,
[tex]80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg[/tex]
The mass of [tex]Mn[/tex] 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 [tex]MnCO_3[/tex] required are, 35 kg
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
