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4 years ago

FeCl3 + NaOH → Fe(OH)3 + NaCl

Chemistry

1 answer:

Sunny_sXe [5.5K]4 years ago

7 0

Answer:

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

Explanation:

You are given a skeleton equation, or unbalanced equation.

FeCl₃ + NaOH → Fe(OH)₃ + NaCl

Count the number of atoms on each side. If there is 1 atom, the subscript is not written. If the subscript to the right of a bracket, it belongs to each element in the bracket.

Reactants → Products

Fe 1 Fe 1

Cl 3 Cl 1 Not balanced

Na 1 Na 1

O 1 O 3 Not balanced

H 1 H 3 Not balanced

The number of each element must be the same in the reactants and products. We can see that Cl (chlorine), O (oxygen) and H (hydrogen) have different numbers.

Write in coefficients in front of formulas (separated by + sign). Coefficients multiply each subscript (or "1" if not written) in the formula. Recount the atoms until you have equal reactant and product atom numbers.

I see that I need more chlorine in my products. Multiply the formula that has "Cl" by 3.

FeCl₃ + NaOH → Fe(OH)₃ + 3NaCl

Reactants → Products

Fe 1 Fe 1

Cl 3 Cl 3 Cl is balanced now.

Na 1 Na 3 Not balanced

O 1 O 3 Not balanced

H 1 H 3 Not balanced

Fix the "Na" now. Multiply the formula with "Na" on the left by 3.

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

Reactants → Products

Fe 1 Fe 1

Cl 3 Cl 3

Na 3 Na 3

O 3 O 3

H 3 H 3

Each side has the same number of each element. Therefore, this is balanced now.

Balanced formula:

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

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3 years ago

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Explanation:

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$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

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Putting values in above equation, we get:

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$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

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Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

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3 years ago

FeCl3 + NaOH → Fe(OH)3 + NaCl​

FeCl3 + NaOH → Fe(OH)3 + NaCl