If 25.6 mL isopropyl alcohol fully decomposes, what mass of H2 is formed? The density of isopropyl alcohol is 0.785 g/mL. g

Answers

Answer 1

Answer:

The correct answer is 0.67 g H₂

Explanation:

Isopropyl alcohol (C₃H₇OH) can decompose to give acetone (C₂H₆OH) and hydrogen gas (H₂) according to the following chemical equation:

C₃H₇OH (g) ⇒ C₂H₆CO(g) + H₂(g)

We can calculate the initial mass of isopropyl alcohol from the density and volume data:

density = m/V = 0.785 g/mL

⇒ m = density x V = 0.785 g/mL x 25.6 mL = 20.096 g C₃H₇OH

According to the chemical equation 1 mol of C₃H₇OH gives 1 mol H₂. The molar mass of C₃H₇OH is:

molar mass C₃H₇OH = (12 g/mol x 3) + (1 g/mol x 7) + 16 g/mol + 1 g/mol = 60 g/mol

molar mass H₂ = 1 g/mol x 2 = 2 g/mol

So, we obtain: 2 g H₂ from 60 g C₃H₇OH. We multiply this stoichiometric ratio (2 g H₂/60 g C₃H₇OH) by the initial mass of C₃H₇OH to obtain the mass of H₂ is formed:

20.096 g C₃H₇OH x (2 g H₂/60 g C₃H₇OH) = 0.6698 g ≅ 0.67 g H₂


Related Questions

Which is an example of a current research focus in chemistry?

A. applying gene therapy to treat certain diseases

B. using hook-and-loop tape in the clothing industry

C. developing smoke detectors for common use

D. studying coal combustion as an energy source

Answers

d if i’m not mistaken

Answer:

b is the correct answer

do not trust answer one

Explanation:

what’s the most abundant isotope of lawrencium

Answers

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .

Answers

Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

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