1. Based on the ideal gas law, there is a simple equivalency that exists between

Question

1. Based on the ideal gas law, there is a simple equivalency that exists between the amount of gas and the volume it occupies. At standard temperature and pressure (STP; 273.15 K and 1 atm, respectively), one mole of gas occupies 22.4 L of volume. What mass of methanol (CH3OH) could you form if you reacted 3.39 L of a gas mixture (at STP) that contains an equal number of carbon monoxide (CO) and hydrogen gas (H2) molecules?

2. Assuming the temperature and volume remain constant, changes to the pressure in the reaction vessel will directly correspond to changes in the number of moles based on the ideal gas law. Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in part A, and the temperature and volume were constant at values of 298 K and 2.00 L, respectively. If the pressure was 8.02 atm prior to the reaction, what would be the expected pressure after the reaction was completed?

3. Hydrogen has also been considered as an alternative fuel for vehicles designed to combust hydrogen and oxygen, which produces water as a product. However, concerns were raised because methane is typically used on a large scale to produce hydrogen gas. Assume that a gallon of gasoline contains 2400 g of carbon. If a gasoline engine achieves 30 miles per gallon, each mile consumes 80 g of carbon (about 107 g of methane contains 80 g of carbon). Alternatively, a hydrogen engine can achieve 80 miles per kilogram of hydrogen gas. What is the mass of methane (CH4) needed to produce enough hydrogen gas (H2) to drive one mile using the theoretical hydrogen engine?

Explanation / Answer

Ans 1:

Since the volume of gas is directly proportional to the number of moles of molecules , so the volume of each gas will be 3.39 /2 = 1.695 L

Number of moles of gas can be determined by the formula , pV = nRT

n = pV / RT

n = ( 1x 1.695) / (0.0821 x 273.15)

n = 0.0756 moles

Now the reaction is :

CO + 2H2 = CH3OH

Since there are equimolar concentrations of both the reactants , hence H2 is the limiting reagent.

So 0.0756 moles of H2 will produce 0.0756 / 2 = 0.0378 moles of CH3OH

Mass of CH3OH = 0.0378 x 32.04

= 1.21 grams

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