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3 years ago

The equation for the synthesis of ammonia is below. How many moles of H 2 are

Chemistry

1 answer:

kkurt [141]3 years ago

4 0

Moles of H₂ are needed to produce 9.33 moles of NH₃ : 13.995

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products

Reaction for the synthesis of ammonia :

N₂+3H₂⇒2NH₃

moles of NH₃ = 9.33

From equation, mol ratio of H₂ : NH₃ = 3 : 2, so mol H₂ :

$\tt =\dfrac{3}{2}\times 9.33\\\\=13.995$

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If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

2H₂ + O₂ -----------> 2H₂O

Now look at the reaction for mole ratio

2H₂ + O₂ -----------> 2H₂O

2 mole 2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

2H₂ + O₂ -----------> 2H₂O

2 mole (2 g/mol) 2 mole (18 g/mol)

4 g 36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

4 g of H₂ ≅ 36 g of H₂O

30 g of H₂ ≅ X of H₂O

Do cross multiplication

X of H₂O = 30 g x 36 g / 4 g

X of H₂O = 270 g

Now convert grams of H₂O into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 270 g / 18 (g/mol)

No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

5 0

3 years ago

Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p

likoan [24]

Answer: 3) 39.96 amu

Explanation:

Mass of isotope Ar- 36 = 35.97 amu

% abundance of isotope Ar- 36= 0.337% = $\frac{0.337}{100}=3.37\times 10^{-3}$

Mass of isotope Ar- 38 = 37.96 amu

% abundance of isotope 2 = 0.063 % = $\frac{0.063}{100}=6.3\times 10^{-4}$

Mass of isotope Ar- 40 = 39.96 amu

% abundance of isotope 2 = 99.600 % = $\frac{99.600}{100}=0.996$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]$

$A=39.96amu$

Therefore, the average atomic mass of argon is 39.96 amu

4 0

4 years ago

How many elements are in the second group? (column)

zavuch27 [327]

Of the periodic table?

6 0

3 years ago

PLEASE HELP ME.!

Karo-lina-s [1.5K]

Answer:

Anaphase II

Explanation:

4 0

3 years ago

Read 2 more answers

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$

$Q=\frac{10^{2} }{0.10^{2} *0.10}$

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

3 0

3 years ago