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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.