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\[2NO(g) + O_2(g) \rightarrow 2NO_2(g)\]

If 98.3 L of \(NO_2\) forms, measured at \(39.0^\circ C\) and \(631 \, \text{mmHg}\), what is the percent yield?

Answer :

The percent yield of the reaction 2NO(g)+O₂(g)→2NO₂(g) is 7.98%.

Mole ratio of NO to NO₂ is 2:2 or 1:1. Therefore, the number of moles of NO₂ formed is equal to the number of moles of NO reacted.

For an ideal gas, PV = nRT

n = PV/RT where temperature is in kelvin

volume of NO₂ formed = 98.3 L

temperature = 39.0°C or 312.15 K

pressure = 631 mmHg or 0.830 atm

n(NO) = PV/RT \

= (0.830 atm)(98.3 L)/(0.0821 L·atm/mol·K)(312.15 K)

= 33.6 mol

Since, the ratio is 1:1 yield of NO₂ is also 33.6 mol.

Actual yield can be found by converting moles in molar mass of NO₂

NO₂: 1 × 14.01 + 2 × 16.00 = 46.01 g/mol

If the mass of NO2 obtained was 123.4 g. Then the number of moles of NO₂ is

n(NO₂) = m/M = 123.4 g/46.01 g/mol = 2.68 mol

% yield = (actual yield/theoretical yield) x 100%

= (2.68 mol/33.6 mol) x 100%

= 7.98%

Learn more about percent yield at:

brainly.com/question/2506978

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