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maxonik [38]
3 years ago
7

Ammonia is produced by the following reaction. 3H2(g) + N2(g)

Chemistry
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

The hydrogen produces the smaller amount of ammonia.  

Step-by-step explanation:

We are given the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      28.02   2.016     17.03

                 N₂  +  3H₂ ⟶ 2NH₃

Mass/g:   70.0    7.00

1. Calculate the moles of N₂ and H₂

Moles N₂ = 70.0 × 1/28.02  

Moles N₂ = 2.498 mol N₂  

Moles H₂ = 7.00 × 2.016  

Moles H₂ = 3.472 mol N₂  

=====

2. Calculate the moles of NH₃ from each reactant

<em>From</em> N₂:

The molar ratio is 2 mol NH₃/1 mol N₂

Moles of NH₃ = 2.498 × 2/1

Moles of NH₃ = 4.996 mol NH₃

<em>From</em> H₂:

The molar ratio is 2 mol NH₃/3 mol H₂  

Moles of NH₃ = 3.472 × 2/3  

Moles of NH₃ = 4.139 mol NH₃

======  

3. Identify the limiting reactant

The limiting reactant is H₂, because it produces fewer moles of NH₃.

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If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation
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Answer:

59.077 kJ/mol.

Explanation:

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where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

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2 years ago
Match each chemical reaction with the type of reaction that best describes it.
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Plugging the given data into the half life equation we have,

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