mistye ChapterSection Maste 5.7.Gas Stoichiometty Google Chio Secure https//sess

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mistye ChapterSection Maste 5.7.Gas Stoichiometty Google Chio Secure https//session.masteringthemistry.com/myct/itemviewtassignmentProblemiD-882381508tof tion stoichomelnc Relatbonships and ine ldeal Gas aw On previous l 3018 l next» mass of methanol Submit My Answers Give Up Part D Hydrogen gas is a very useful reagent with many uses in the petroleum, food, and chemical industries. Most hydrogen exists in covalently bonded molecules, and atmospheric air contains less than 1 ppm of diatomic hydrogen. Therefore, hydrogen gas is produced on a large scale for these uses, where stea The more economic reaction of steam reforming is the reverse of the reaction depicted in Carbon Monoxide and Hydrogen Sample 1 in the Simulation. m reforming with methane and electrolysis of water are two of the primary method 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 galton 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 ve 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? Express the mass in grams to two significant digits. + Hints 01| mass of methane (CH 3:30 PM ×10/17/2017

Explanation / Answer

part C

CO + 2H2 ---> CH3OH

equal number of molecules means equal no of moles.

No of mol of gas present in mixture = PV/RT

         = 1*7.37/(0.0821*273.15)

         = 0.329 mol

Both CO,H2 in equal ratio

No of mol of CO gas = 0.329/2 = 0.1645 mol

No of mol of H2 gas = 0.1645 mol

Limiting reactant = H2

No of mol of cH3oh formed = 0.1645*1/2 = 0.08225 mol

mass of cH3oh formed = 0.08225*32 = 2.63 g

part D

CH4 + H2O ---> 3H2 + CO

hydrogen engine = 80 mile/kg

1 mile = 1/80 = 0.0125 kg H2 gas required

mass of H2 required = 0.0125 kg = 12.5 g

noof mole of H2 consumed per mile = 12.5/2 = 6.25 mol

from equation, 1 mol CH4 = 3 mol H2

so that,

no of mol of CH4 required = 6.25/3 = 2.0833 mol

mass of CH4 required = 2.0833*16 = 33 g

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