Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g_ + 2H2 (g) → CH3OH (g) A 1.50L 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 397 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams.

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 2 mole of react to give 1 mole of

So, 0.0601 moles of react with moles of

Now we have to calculate the mass of

Therefore, the theoretical yield of methanol is, 0.96 grams.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-10-22T07:09:17+02:00

The theoretical yield of methanol is 0.496 g of methanol.

The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).

From the partial pressures of each reactant, we can obtain the number of moles of reactants.

For CO;

P = 232 mmHg or 0.305 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.018 moles

For hydrogen;

P = 397 mmHg or 0.522 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.031 moles

From the reaction equation;

1 mole of CO reacted with 2 moles of H2

0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole

= 0.036 moles of H2

We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.

2 moles of H2 yields 1 mole of methanol

0.031 moles of H2 yields 0.031 moles × 1 moles/2 mole

= 0.0155 moles of methanol

Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol

= 0.496 g of methanol

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