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2 years ago

12

How many moles of hydrogen are needed to produce 13.78 mol of ethane?

2 answers:

mamaluj [8]2 years ago

4 0

Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.

Likurg_2 [28]2 years ago

4 0

This is an incomplete question, here is a complete question.

The following balanced equation shows the formation of ethane (C₂H₆).

$C_2H_2+2H_2\rightarrow C_2H_6$

How many moles of hydrogen are needed to produce 13.78 mol of ethane?

1) 3.445 mol

2) 6.890 mol

3) 27.56 mol

4) 55.12 mol

Answer : The number of moles of hydrogen needed are, (3) 27.56 mol

Explanation : Given,

Moles of ethane = 13.78 mol

The given balanced chemical reaction is:

$C_2H_2+2H_2\rightarrow C_2H_6$

From the balanced chemical reaction we conclude that,

As, 1 mole of ethane produced from 2 moles of hydrogen

So, 13.78 mole of ethane produced from $13.78\times 2=27.56$ moles of hydrogen

Thus, the number of moles of hydrogen needed are, 27.56 mol

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How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

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2 years ago

Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change

yuradex [85]

Answer:

ΔH°r = -184.6 kJ

Explanation:

Let's consider the following balanced equation.

H₂(g) + Cl₂(g) ⇄ 2 HCl(g)

We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression:

ΔH°r = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(p) are the standard enthalpies of formation of reactants and products

By definition, the standard enthalpy of formation of a simple substance in its most stable state is zero. Then,

ΔH°r = 2 mol × ΔH°f(HCl(g)) - [1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(Cl₂(g))]

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5 0

2 years ago

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

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now find out number of moles (n)

n = mass / molar mass

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n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

2 years ago