Question

You might be able to roughly estimate mass in a visual manner, but it could be easy to dismiss the fact that a mole or even a gram represents a significantly large number of molecules. Avogadro's number, 6.022×1023, describes how many molecules, atoms, or particles are in one mole of a substance. We use moles to describe and compare amounts because the values correspond to the macroscopic scale in which we operate. Let us consider a reaction that takes place between relatively low masses. In Sample 2 from the simulation, carbon monoxide gas (CO) can react with hydrogen gas (H2) to form methanol (CH3OH), which is a liquid at room temperature. If you had a sealed reaction vessel that was vacuum pumped, and measured its change in mass as you introduced each reactant gas to the vessel, you could determine the amount of each gas that entered the vessel by those measured changes in mass. If 1.50 μg of CO and 6.80 μg of H2 were added to a reaction vessel, how many total gas particles would remain in the reaction vessel after the reaction went to completion? Assume no gas particles dissolve into the methanol.

Answer 1

The balanced equation is

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

Which means, every 2 moles of H2 will react with 1 mole of CO.

Find the number of moles, n = m/Mr

CO = (1.5x10^-6)/(12) = 1.25 x 10^-7 mol

H2 = (6.8x10^-6)/(2) = 3.4 x 10^-6 mol

Since we have a mole ratio of 2 to 1, H2 value will be multipled by 2 as 2 moles of H2 react to 1 mole of CO.

3.4 x 10^-6 x 2 = 6.8 x 10^-6

So obviously we can see that H2 is in excess and only some H2 molecules will be left at the end of the reaction.

6.8 x 10^-6 - 1.25 x 10^-7 = 6.675 x 10^-6 Moles of unreacted H2

To find the molecules of this amount of moles, Multiply it by Avogadros number, 6.02 x 10^23

Because for every mole, there are 6.02 x 10^23 molecules

6.575 x 10^-6 x 6.02 x 10^23 = 3.96 x 10^ 18 Molecules of gas

Answer 2

Answer:

1.98*10^18 particles of H2

Explanation:

The balanced equation is

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

The number of moles of each component is calculated as follows:

moles = mass/molecular weight

where mass is in grams and molecular weight in grams/mol

Then:

moles of CO = (1.5x10^-6)/(28) = 5.35*10^-8 mol

moles of H2 = (6.8x10^-6)/(2) = 3.4*10^-6 mol

From the balanced equation we now that 2 moles of H2 will react with 1 mole of CO. Then, if 5.35*10^-8 mol of CO2 reacts, the following proportion must be satisfied:

2 moles of H2/x moles of H2  = 1 mole of CO/5.35*10^-8 mol of CO2

x = 5.35*10^-8/1 * 2 = 1.07*10^-7 moles of H2

But 3.4*10^-6 mol of H2 entered to the vessel, so 3.4*10^-6 - 1.07*10^-7 = 3.293*10^-6 moles of H2 will left after the completion of the reaction (all CO reacted). Using Avogadro's number we can calculate the number of particles would remain in the reaction vessel:

3.293*10^-6 mole of H2 * 6.022×10^23 particles/mole = 1.98*10^18 particles of H2