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%
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