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3 years ago

Iron react with oxygen to form Iron (III) Oxide. How many moles of iron (III) Oxide is in 138.5g?

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Answer:

0.87mole

Explanation:

The reaction equation is given as;

4Fe + 3O₂ → 2Fe₂O₃

Number of moles of Fe₂O₃ in 138.5g

To solve this problem;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Fe₂O₃ = 2(56) + 3(16) = 160g/mol

Now;

Number of moles = $\frac{138.5}{160}$ = 0.87mole

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Read 2 more answers

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and

n200080 [17]

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

<em>Where [] are concentrations at equilibrium</em>

And Q is:

Q = [HI]² / [H2] [I2]

<em>Where [] are actual concentrations of the reactants.</em>

<em />

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

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