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How many liters of gaseous oxygen are needed to produce 4.50 L of gaseous nitrogen dioxide, if both gases are being measured at STP?

User Temaruk
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1 Answer

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answer and explanation

the balanced chemical reaction for the formation of nitrogen dioxie is

2NO + O₂ → 2NO₂

and so at STP the number of mols of nitrogen dioxide in 4.50L would be

PV = nRT

n = PV/RT

= (1atmx 4.50L) / (8.314 x 273.15K

= 4.50/2270.97

= 0.00204 mols

0.00204 mols of nitrogen dioxide would have been formed from the reaction.

we know from the balance reaction that 1 mol of oxygen forms 2 mols of nitrogen dioxide and so the mols of oxygen that were present would be

1/2 x0.00204 = 0.00102 mols

we know that 1 mol of a gas occupies a volume of 22.4L at STP and so

0.00102 mols of oxygen would have a volume of

0.0102 mols x 22.4 L/mol

= 0.0228L

0.0228L of oxygen would be needed

User ForgetfulFellow
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