Answer:
1) The limiting reactant is: NO.
2) The no. of moles of NOâ‚‚ 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 Oâ‚‚(g) to produce 2.0 moles NOâ‚‚(g).
NO reacts with Oâ‚‚ with (2: 1) molar ratio.
So, 0.863 mol of NO reacts completely with 0.4315 mol of Oâ‚‚ with the stechiometric molar ratio (2: 1) and the excess Oâ‚‚ is (0.501 mol - 0.4315 mol = 0.0695 mol).
So:
The limiting reactant is: NO.
2) To get the no. of NOâ‚‚ 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 NOâ‚‚ produced = 0.863 mol.
A. The limiting reactant in the reaction is NO
B. The number of mole of NOâ‚‚ 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 Oâ‚‚.
Therefore,
0.863 mole of NO will require = 0.863 / 2 = 0.4315 mole of Oâ‚‚
From the above calculation, we can see that only 0.4315 mole out of 0.501 mole of Oâ‚‚ 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 NOâ‚‚.
Therefore,
0.863 mole of NO will also react to produce 0.863 mole of NOâ‚‚.
Thus, 0.863 mole of NOâ‚‚ was obtained from the reaction.
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