Chemistry, 03.08.2019 00:30 trinigal83

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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Answer from: Lizae9099

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 from: sjsmith21

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, 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 .

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 = 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

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

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 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

Answer from: Quest

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.