This is an example of what process ?

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

Answer: The mass of decane produced is $1.743\times 10^2g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

3 years ago

How many total electrons can the p orbitals hold? A. 6 B. 2 C. 10 D. 3

mars1129 [50]

Answer:

A. 6

Explanation:

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8 0

3 years ago

Read 2 more answers

Select True or False: Hydrogen plays an important role in many industrial processes. The following correctly represents a balanc

Orlov [11]

Answer: False

Explanation:

Since the given equation is not balanced properly.

Since oxygen and hydrogen atoms are not balanced.

There should be 6 H2O (g) molecules and 14 mol H2 (g)

5 0

3 years ago

T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,

azamat

Answer:

TRUE
FALSE
TRUE
FALSE
FALSE
TRUE

Explanation:

At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
Faster solidification results in smaller interlamellar spacing : TRUE

4 0

3 years ago