Question

A substance used as a cleaner and a fuel is 37.48% C, 49.93% O and 12.58% H by mass. A 0.2804-g sample of the substance occupies a volume of 250.0 mL when it is vaporized at 75o C and 1.00 atm of pressure.
R = 0.0821 L atm/ mol K
a) This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas?
b) If 59.6 kg of the product is actually produced, given the reaction described in (a), what is the percent yield?
c) This compound (the substance you identified in part a) is a potential replacement for gasoline. The products of the complete combustion of this fuel are the same as those for the complete combustion of a hydrocarbon (CO2 and H2O). Calculate the volume of CO2 produced at 27o C and 766 mmHg when 1.00 gallon of this fuel is completely combusted. The density of the fuel is 0.7914 g/mL. 1 gallon = 3.785 liters
d) A claim was made that this fuel is better for the environment because it produces less CO2 per gallon than gasoline, which can be represented by the formula C8H18 (octane). Is this claim true? Octane has a density of 0.6986 g/mL

Answer 1

Answer:

Explanation:

We shall find out the molecular formula of the substance .

Ration of number of atoms of C , O and H

= $\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}$

= 3.12 : 3.12 : 12.58

= 1 : 1 : 4

volume of gas at NTP

= 250 x 273 / 350 mL .

= 195 mL .

Molecular weight of the substance = .2804 x 22400 / 195 g

= 32. approx

Let the molecular formula be

(COH₄)n  

n x 32 = 32

n = 1

Molecular formula = COH₄

The compound appears to be CH₃OH

a )

CO + 2H₂ = CH₃OH

28g     4g          32g

59      8

For 8 kg hydrogen , CO required = 56 kg

CO is in excess .  hydrogen is the limiting reagent .

mass of product formed

= 32 x 8 / 4

= 64 kg

b )

percentage yield = product actually formed / product to be formed theoretically  x 100

= 59.6 x 100 / 64

= 93.12 %

c )

2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .

64 g                     2 x 22.4 L

Gram of gas in 1 gallon of fuel

= .7914 x 3785

= 2995.5 g

CO₂ produced at NTP by 2995.5 g CH₃OH

= 2 x 22.4 x 2995.5 / 64 L

= 2096.85 L

At 27° C and 766 mm Hg , this volume is equal to

2096.85 x 300 x 760 / 273 x 766

= 2286.18  L .

d )

C₈H₁₈  =  8CO₂

114g           8 x 22.4 L

gram of fuel per unit gallon

= .6986 x 3785

= 2644.2g

gram of CO₂ produced by 1 gallon of fuel  at NTP

= 8 x 22.4 x 2644.2 / 114

= 4156.5 L

So it produces more CO₂ .