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Luda [366]
3 years ago
14

When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because:

Chemistry
1 answer:
noname [10]3 years ago
6 0

7.5 mol of hydrogen would be needed to consume the available nitrogen.

Explanation:

When hydrogen reacts with nitrogen, ammonia is formed as shown below;

3H₂ (g) + N₂ (g) → 2NH₃ (g)

As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.

The limiting factor in a chemical reaction is the reactant that gets depleted first.

Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.

The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.

However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5

3.5 moles fall short of the 7.5 moles of H₂ required for a complete reaction. H₂ gets depleted first before N₂.  The reaction would require 4 more moles of H₂.

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When a physical change in a sample occurs, which of the following does NOT change?
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How many grams of salt are in 1 kilogram of seawater with average salinity? quizzlet?
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3 years ago
The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

6 0
3 years ago
23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
3 years ago
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