At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound Amount

$Fe_2O_3$ 3.54 g

$H_2$ 3.63 g

Fe 2.37 g

$H_2O$ 2.13 g

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

Answer: The value of equilibrium constant for the given reaction is $2.8\times 10^{-4}$

Explanation:

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

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

For hydrogen gas:

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of hydrogen gas}=\frac{3.63}{2\times 5.4}\\\\\text{Molarity of hydrogen gas}=0.336M$

For water:

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of water}=\frac{2.13}{18\times 5.4}\\\\\text{Molarity of water}=0.0219M$

The given chemical equation follows:

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

The expression of $K_c$ for above equation follows:

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

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

$K_c=\frac{(0.0219)^3}{(0.336)^3}\\\\K_c=2.77\times 10^{-4}$

Hence, the value of equilibrium constant for the given reaction is $2.8\times 10^{-4}$

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3 years ago

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How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution r

NNADVOKAT [17]

2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃

m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol

1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw

2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}

m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}

m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}

m(Fe)=55.847*245*0.805/179.857= 61.24 g

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3 years ago

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