Answer:
1) The limiting reactant is: NO.
2) The no. of moles of NO₂ produced = 0.863 mol.
Explanation:
1)
From the balanced reaction:
2NO(g) + O₂(g) → 2NO₂(g).
It is clear that 2.0 moles of NO(g) react with 1.0 mole of O₂(g) to produce 2.0 moles NO₂(g).
NO reacts with O₂ with (2: 1) molar ratio.
So, 0.863 mol of NO reacts completely with 0.4315 mol of O₂ with the stechiometric molar ratio (2: 1) and the excess O₂ is (0.501 mol - 0.4315 mol = 0.0695 mol).
So:
The limiting reactant is: NO.
2) To get the no. of NO₂ produced, we use cross multiplication:
2.0 moles of NO(g) → produce 2.0 moles NO₂(g), from the sytichiometry.
∴ 0.863 mol of NO(g) → produce 0.863 moles NO₂(g).
So, The no. of moles of NO₂ produced = 0.863 mol.
A. The limiting reactant in the reaction is NO
B. The number of mole of NO₂ produced from the reaction is 0.863 mole
Data obtained from the question:
Balanced equation
2NO(g) + O₂(g) → 2NO₂(g)
From the balanced equation above,
2 moles of NO required 1 mole of O₂.
Therefore,
0.863 mole of NO will require = 0.863 / 2 = 0.4315 mole of O₂
From the above calculation, we can see that only 0.4315 mole out of 0.501 mole of O₂ given, is required to react completely with 0.863 mole of NO.
Therefore, NO is the limiting reactant
The limiting reactant will be used in this case
2NO(g) + O₂(g) → 2NO₂(g)
From the balanced equation above,
2 moles of NO reacted to produce 2 moles of NO₂.
Therefore,
0.863 mole of NO will also react to produce 0.863 mole of NO₂.
Thus, 0.863 mole of NO₂ was obtained from the reaction.
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