9 L of the NO2 was produced if 9.0 L of NO are reacted with excess at STP.
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
According to the question, the given reaction is -
2NO(g) + O₂(g)--->2NO₂(g)
Since at STP one mole of a gas occupies the volume of 22.4 L.
As given in the question -
9 L of NO, i.e .
22.4 L = 1 mol
1 L =
For 9 L = = 0.40 mol
From the chemical reaction,
The Oxygen is in excess, hence NO becomes the limiting reagent, and will determine the moles of the product.
Hence, 2 moles of NO will produce 2 moles of NO₂.
Therefore, 0.40 mol of NO will produce 0.40 mol of NO₂.
Hence, the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 X 22.4 L = 9 L
Hence, 9 L of the NO₂ was produced.
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