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kirza4 [7]
3 years ago
12

Balance FeCl3 + MgO ---> Fe2O3 + MgCl2

Chemistry
1 answer:
german3 years ago
3 0

Answer: The balanced equation is 2FeCl_{3} + 3MgO \rightarrow Fe_{2}O_{3} + 3MgCl_{2}

Explanation:

The given reaction equation is as follows.

FeCl_{3} + MgO \rightarrow Fe_{2}O_{3} + MgCl_{2}

Here, number of atoms present on reactant side are as follows.

  • Fe = 1
  • Cl = 3
  • Mg = 1
  • O = 1

Number of atoms on product side are as follows.

  • Fe = 2
  • Cl = 2
  • Mg = 1
  • O = 3

To balance this equation, multiply FeCl_{3} by 2 and MgO by 3 on reactant side. Also, multiply MgCl_{2} by 3 on product side. Therefore, the equation can be rewritten as follows.

2FeCl_{3} + 3MgO \rightarrow Fe_{2}O_{3} + 3MgCl_{2}

Hence, number of atoms on reactant side are as follows.

  • Fe = 2
  • Cl = 6
  • Mg = 3
  • O = 3

Number of atoms on product side are as follows.

  • Fe = 2
  • Cl = 6
  • Mg = 3
  • O = 3

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that the balanced equation is 2FeCl_{3} + 3MgO \rightarrow Fe_{2}O_{3} + 3MgCl_{2}.

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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.

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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.

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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

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3 years ago
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3. How many grams of water will be produced from 15.0 grams of Methane? *<br> CH4 +2 02 → CO2 + 2H2O
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16 gm CH4 + 64 gm O2 = 44 gm CO2 + 36 gm H2O

Since 16 gm CH4 produce 36 gm H2O

Hence 2.5 gmCH4 produce 36×2.5/16 gm H2O

= 5.265 gm of H2O
2.1K viewsView 2 Upvoters

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