Question

1.Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction:

CO(g)+2H2(g)→CH3OH(g)CO(g)+2H2(g)→CH3OH(g)

A 1.65 LL reaction vessel, initially at 305 KK, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 395 mmHg.

Identify the limiting reactant and determine the theoretical yield of methanol in grams.

Express your answer with the appropriate units.

2.

What is the pressure in a 12.0-LL cylinder filled with 39.5 g of oxygen gas at a temperature of 328 K?

Express your answer to three significant figures with the appropriate units.

3.A 275-mL flask contains pure helium at a pressure of 743 torr . A second flask with a volume of 470 mL contains pure argon at a pressure of 731 torr.

part a:

If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of helium?

Express the partial pressure in torr to three significant figures.

part b:

If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of argon?

Express the partial pressure in torr to three significant figures.

Part C:

If we connect the two flasks through a stopcock and we open the stopcock, what is the total pressure?

Express the total pressure in torr to three significant figures.

4.

A sample of gas has a mass of 38.6 mg. Its volume is 228 mL at a temperature of 54 ∘C∘C and a pressure of 842 torr. Find the molar mass of the gas.

Express your answer in grams per mole to three significant figures.

Answer 1

Solution (1) :

CO(g)+2H2(g) ------ > CH3OH(g)

Given,

Volume = 1.65 L

Temperature = 305 K

Pressure = (232 + 395 mmHg )* 1 atm /760 mmHg = 0.825 atm

First we find out the total moles of the reactants in the vessel using this formula :

PV= nRT

Or, PV/RT = n

(0.825 atm * 1.65 L) / (0.08206 L atm per mol K * 305 K) = n

on solving we will get, n = 0.0543 moles

Now lets calculate the moles of each gas using the partial pressures

Total pressure = 232 mmHg + 395 mmHg = 627 mmHg

Moles of CO = (partial pressure / total pressure ) X total moles

= (232 mmHg / 627 mmHg) X 0.0543 mol

= 0.02 mol CO

Moles of H2 = total moles - moles of CO

= 0.0543 - 0.02 mol H2

=0.0343 mol H2

Now lets calculate the moles of the CH3OH that can be oroduced from each reactant using the mole ratio of the each reactant

(0.02 mol CO * 1 mol CH3OH / 1 mol CO) = 0.02 mol CH3OH

(0.0343 mol H2 * 1 mol CH3OH / 2 mol H2)= 0.01715 mol CH3OH

H2 gives less moles of product

Therefore H2 is the limiting reactant

So the moles of the CH3OH that can be formed are 0.01715 mol CH3OH

Now lets calculate the mass of CH3OH

Mass= moles * molar mass

= 0.01715 mol CH3OH * 32.04 g per mol

= 0.549 g CH3OH

So the theoretical mass of CH3OH that can be formed = 0.549 g