It has been proposed that methanol, CH 3 OH, beconsidered as the replacement for

Question

It has been proposed that methanol, CH3OH, beconsidered as the replacement for gasoline as the major fuel fortransportation in this country because of its high combustionefficiency. Another reason is that methanol can be made easily(cheaply) from heating the coal with steam passing at hightemperature: CO(g) + 2H2(g) -->CH3OH(l) (A) Which is the limiting reagent when 74.5 gramsof CO is mixed with 12.0 grams of H2? (B) After the reaction is completed, how much ofwhich reactant will be left? (C) How much methanol (in grams) should be formedafter the reaction is completed? (D) If 80.3 grams of methanol is actuallyproduced, calculate the product yeild. (E) The density of methanol is 0.7914 g/mL.Calculate the volume of methanol (in gallons) producedin answer (D) above. A small car of of 4- cylinderengine averages 24 miles per gallon in typical city driving. Howfar can this small car go with 80.3 grams of methanol (from(D))? It has been proposed that methanol, CH3OH, beconsidered as the replacement for gasoline as the major fuel fortransportation in this country because of its high combustionefficiency. Another reason is that methanol can be made easily(cheaply) from heating the coal with steam passing at hightemperature: CO(g) + 2H2(g) -->CH3OH(l) (A) Which is the limiting reagent when 74.5 gramsof CO is mixed with 12.0 grams of H2? (B) After the reaction is completed, how much ofwhich reactant will be left? (C) How much methanol (in grams) should be formedafter the reaction is completed? (D) If 80.3 grams of methanol is actuallyproduced, calculate the product yeild. (E) The density of methanol is 0.7914 g/mL.Calculate the volume of methanol (in gallons) producedin answer (D) above. A small car of of 4- cylinderengine averages 24 miles per gallon in typical city driving. Howfar can this small car go with 80.3 grams of methanol (from(D))?

Explanation / Answer

CO(g) + 2H2(g) -->CH3OH(l) Molar mass of CO is = 12 + 16 = 28 g Molar mass of CH3OH is = 12 + 3 * 1 + 16 + 1 = 32g Molar mass of H2 is = 2 * 1 = 2 g 28 g of CO reacts with 2 * 2 (=4 g ) of H2 74.5 g of CO reacts with X g ofH2 X = ( 74.5 * 4 ) / 28       = 10.6428 g ofH2 So , 12 - 10.6428 g of H2 is left unreacted.Sincewhich is the excess reagent. In this case CO is the limiting reagent.                --------------(A) The excess reagent is H2 & the reactant left= 12 - 10.6428 = 1.3572 g of H2      -------(B) 28 g of CO upon reaction produces produces 32 gof CH3OH 74.5 g of CO upon reaction producesproduces Y g of CH3OH Y = ( 74.5 * 32 ) / 28     = 85.1428 g of CH3OH       -----------(C) Percentage of yield = ( actual yield /Theoritical yield ) * 100                              = 94.3 %     -----------(D) Density = mass / Volume

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