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3 years ago

Balance the following equations: HELP ME PLEASE!

Chemistry

1 answer:

noname [10]3 years ago

5 0

Answer:

Explanation:

Balance Chemical Equation:

A balanced chemical equation is that in which the number of the reactant atoms equal to the number of the product atom.

For example if the oxygen at the reactant side is 4, the number of oxygen must be 4 on the product side.

Solution

1. Au₂S₃ + H₂ --------> Au + H₂S

Balance Chemical Equation

1 Au₂S₃ + 3 H₂ --------> 2 Au + 3 H₂S

2. FeS + O₂ ---------> Fe₂O₃ + SO₂

Balance Chemical Equation

4FeS + 7O₂ ---------> 2Fe₂O₃ + 4SO₂

3. NH₃ + O₂ ---------> NO + H₂O

Balance Chemical Equation

4 NH₃ + 5 O₂ ---------> 4 NO + 6 H₂O

4. BaO + H₂O --------> Ba(OH)₂

Balance Chemical Equation

1 BaO + 1 H₂O --------> 1 Ba(OH)₂

Now by balancing all the atoms in equation are equal in reactant and product

Details also in attachment.

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Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

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

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

<u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

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