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Rudik [331]
3 years ago
14

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide: 2 MnCO3(s) + O2(g) → 2MnO2(s) + 2CO2(g) In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide: 3 MnO2(s) + 4 Al(s) → 3 Mn(s) + 2 Al2O3(s) Write the net chemical equation for the production of manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.
Chemistry
1 answer:
hjlf3 years ago
5 0

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

Explanation :

The given two chemical reactions are:

(1) 2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

(2) 3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) 6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)

Now we are adding both the reactions, we get the overall chemical reaction.

6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)

The  MnO_2 is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

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Answer:

m_{Cr_2O_3}^{actual}=62.4gCr_2O_3

Explanation:

Hello!

In this case, according to the reaction:

4Cr+3O_2\rightarrow 2Cr_2O_3

We can see there is a 4:2 mole ratio between chromium and chromium (III) oxide, this, for the given 56.2 g of chromium, the theoretical yield of the oxide product is computed down below:

m_{Cr_2O_3}^{theoretical}=56.2gCr*\frac{1molCr}{52.0gCr}*\frac{2molCr_2O_3}{4molCr} *\frac{151.99gCr_2O_3}{1molCr_2O_3}  =82.13gCr_2O_3

Now, considering the 76.0-% yield for this reaction, the actual yield turns out:

m_{Cr_2O_3}^{actual}=82.13gCr_2O_3*\frac{76.0gCr_2O_3}{100gCr_2O_3} \\\\m_{Cr_2O_3}^{actual}=62.4gCr_2O_3

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3 years ago
HELP!! b. At the equivalence point, all of the acid has been neutralized by the base. Why does the pH change so
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kupik [55]

Answer:

5446.8 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 50 g

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Specific heat capacity (C) = 0.89 J/gºC

Heat (Q) required =?

Next, we shall determine the change in the temperature. This can be obtained as follow:

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 192.4 – 70

ΔT = 122.4 °C

Finally, we shall determine the heat required to heat up the block of aluminum as follow:

Mass (M) = 50 g

Specific heat capacity (C) = 0.89 J/gºC

Change in temperature (ΔT) = 122.4 °C

Heat (Q) required =?

Q = MCΔT

Q = 50 × 0.89 × 122.4

Q = 5446.8 J

Thus, the heat required to heat up the block of aluminum is 5446.8 J

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