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77julia77 [94]
3 years ago
10

• Balance this chemical equation.

Chemistry
1 answer:
user100 [1]3 years ago
4 0

Answer:

Explanation:

Chemical equation:

FeCl₃  +  NaOH   →  Fe(OH)₃ + NaCl

Balanced chemical equation:

FeCl₃  +  3NaOH   →  Fe(OH)₃ + 3NaCl

Step 1:

FeCl₃  +  NaOH   →  Fe(OH)₃ + NaCl

Fe = 1                             Fe = 1

Cl =  3                            Cl = 1

Na =  1                            Na =  1

OH = 1                             OH = 3

Step 2:

FeCl₃  +  3NaOH   →  Fe(OH)₃ + NaCl

Fe = 1                              Fe = 1

Cl =  3                             Cl = 1

Na =  3                            Na =  1

OH = 3                             OH = 3

Step 3:

FeCl₃  +  3NaOH   →  Fe(OH)₃ + 3NaCl

Fe = 1                              Fe = 1

Cl =  3                             Cl = 3

Na =  3                            Na =  3

OH = 3                             OH = 3

Now we the equation is completely balanced and follow  the law of conservation of mass because there are equal number of atoms of each elements are present on both side.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction .

Type of reaction:

It is double displacement reaction.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

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One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

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From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg

Now we have to calculate the percent purity of Fe_2O_3 in the original sample.

Mass of original sample = 2.86\times 10^3kg

\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100

\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

Therefore, the percent purity of Fe_2O_3 in the original sample is 87.94 %

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