Carbon monoxide gas reacts with hydrogen gas to form methanol. CO(g)+2H2(g)→CH3OH(g) A 1.45 L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 353 mmHg .
Identify the limiting reactant and determine the theoretical yield of methanol in grams.
Express your answer with the appropriate units.
pressure of CO = 232 / 232 + 353
= 0.4 mmHg
partial pressure of H2 = 353 / 232+ 353
= 0.6 mmHg
now calculate the no of moles using PV = nRT
no of moles of CO
0.4 x 1.45 = n x 0.0821 x 305
0.58 = n x 25.04
n = 0.0232 moles
similarly no of moles of H2
0.6 x 1.45 = n x 0.0821 x 305
n = 0.87 / 25.04
= 0.035 moles
according to given balanced reaction one mole of CO required 2 moles of H2
that means 0.0232 moles CO required 2 x 0.0232 moles H2 = 0.046
but we have 0.035 moles that means H2 is limiting agent
from the equation from 2 moles of H2 one mole of CH3OH will form so
from 0.035 moles 0.035/2 moles of MeOH will form = 0.0175 moles
grams of MeOH = no of moles x molar mass
= 0.0175 x 32.04
= 0.56 grams
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