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alexandr402 [8]
3 years ago
14

Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi

ng to the following equation:
3H2(g) + N2(g) → 2NH3(g)

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
L

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
L
Chemistry
2 answers:
Westkost [7]3 years ago
6 0
1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
 

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol =  84 liters of H2

Answer: 84 liters

 </span>



vaieri [72.5K]3 years ago
4 0

the answer is

125 & 84

the second part is

0.38

0.38

27

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Molodets [167]
Molar solubility<span> is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

</span>3Cu2+ + 2(AsO4)3-<span> = Cu3(AsO4)2
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7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M
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ivanzaharov [21]
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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
If the decomposition of a sample of k c l o 3 kclox3 produces 3. 29 g of o 2 ox2 , what was the mass (g) of the original sample?
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The mass (g) of the original sample after decomposition is 8.3983 g.

A decomposition reaction can be described as a chemical reaction wherein one reactant breaks down into or extra merchandise.

explanation:

Reaction          2KClO₃ ⇒ 2KCl + 3O₂

moles               2               2            3

molar mass    122.55        74.55    32

Given, Mass of O₂ = 3.29g ⇒ moles of O₂

                                            =  (3.29/32) = 0.1028

3 moles of O₂  produced by 2 moles of KClO₃

Therefore, 0.1028 moles of O₂  produced by (2*0.1028/3) = 0.06853 moles of Kclo₃

Mass of KClo₃ in original sample is = moles * molar mass

                                                        = 0.06853 * 122.55

                                                       = 8.3983 g

A decomposition response occurs whilst one reactant breaks down into or extra merchandise. this may be represented through the general equation: XY → X+ Y. Examples of decomposition reactions consist of the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.

Learn more about decomposition here:-brainly.com/question/27300160

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5 0
1 year ago
If you stacked Avogadro's number of pennies one on top of the other on Earth's surface, how far would the stack extend (in kilom
attashe74 [19]

Answer:

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

<em>Note: The question is missing some parts. The complete question is as follows;</em>

<em>A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]</em>

Explanation:

Avogadro number = 6.02 * 10²³

thickness of a penny = 1.0 mm

I mm = 0.001 m

Thickness of Avogadro number of pennies stacked one upon another will be:

6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m

Distance in km;

1 m = 0.001 km

therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

5 0
3 years ago
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