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If 6.50 L of water vapor at 50.2 °C and 0.121 atm reacts with excess iron, how many grams of iron(III) oxide will be produced?

2Fe(s)+3H2O(g)⟶Fe2O3(s)+3H2(g)

Sagot :

Eduard22slyidn

Answer:

1.60 g of Fe₂O₃

Explanation:

We'll begin by calculating the number of mole water that reacted. This can be obtained as follow:

Volume (V) = 6.50 L

Temperature (T) = 50.2 °C = 50.2 + 273 = 323.2 K

Pressure (P) = 0.121 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.121 × 6.5 = n × 0.0821 × 323.2

0.7865 = n × 26.53472

Divide both side by 26.53472

n = 0.7865 / 26.53472

n = 0.03 mole

Thus, 0.03 mole of water reacted.

Next, we shall determine the number of mole of Fe₂O₃ produced from the reaction. This can be obtained as follow:

2Fe + 3H₂O —> Fe₂O₃ + 3H₂

From the balanced equation above,

3 moles of H₂O reacted to produce 1 mole Fe₂O₃.

Therefore, 0.03 mole of H₂O will react to produce = (0.03 × 1)/3 = 0.01 mole of Fe₂O₃.

Thus, 0.01 mole of Fe₂O₃ was produced from the reaction.

Finally, we shall determine the mass of 0.01 mole of Fe₂O₃. This can be obtained as follow:

Mole of Fe₂O₃ = 0.01 mole

Molar mass of Fe₂O₃ = (56×2) + (16×3)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ =?

Mass = mole × molar mass

Mass of Fe₂O₃ = 0.01 × 160

Mass of Fe₂O₃ = 1.60 g

Therefore, 1.60 g of Fe₂O₃ were produced.

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