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scarlettp13
03.08.2019 •
Chemistry
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step 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.
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Ответ:
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
Ответ:
Answer : The mass of
required are, 35 kg
Explanation :
First we have to calculate the mass of
.
The first step balanced chemical reaction is:
Molar mass of
= 115 g/mole
Molar mass of
= 87 g/mole
Let the mass of
be, 'x' grams.
From the balanced reaction, we conclude that
As,
of
react to give
of ![MnO_2](/tpl/images/0163/6595/00c21.png)
So,
of
react to give
of ![MnO_2](/tpl/images/0163/6595/00c21.png)
And as we are given that the yield produced from the first step is, 65 % that means,
The mass of
obtained = 0.4542x g
Now we have to calculate the mass of
.
The second step balanced chemical reaction is:
Molar mass of
= 87 g/mole
Molar mass of
= 55 g/mole
From the balanced reaction, we conclude that
As,
of
react to give
of ![Mn](/tpl/images/0163/6595/042d1.png)
So,
of
react to give
of ![Mn](/tpl/images/0163/6595/042d1.png)
And as we are given that the yield produced from the second step is, 80 % that means,
The mass of
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
required are, 35 kg
Ответ: