Nitric oxide (no) reacts with oxygen gas to form nitrogen dioxide (no2), a dark brown gas: 2no(g) + o2(g) → 2no2(g) in one experiment, 0.863 mol of no is mixed with 0.501 mol of o2. determine which of the two reactants is the limiting reactant. calculate also the number of moles of no2 produced. limiting reactant: moles of no2 produced: moles

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.

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