Nitrogen and hydrogen react to form ammonia. Consider the mixture of N2 (blue spheres) and H2 (gray spheres) in the illustration
. Which statement concerning the reaction is MOST accurate?
1 answer:
The illustration is attached below and Options are as follow,
<span>A) Nitrogen is in excess.
B) Hydrogen is in excess.
C) This is a synthesis reaction.
D) Hydrogen is the limiting reactant.
Answer:
Option-D (Hydrogen is the limiting reactant).
Explanation:
The equation of reaction is as follow,
N</span>₂ + 3 H₂ → 3 NH₂
According to this equation 1 mole of nitrogen reacts with 3 moles of hydrogen gas to produce 3 moles of ammonia.
In given illustration 4 moles of N₂ (blue spheres) are present and 6 moles of H₂ are present.
As,
1 mole of N₂ requires = 3 moles of H₂
So,
4 moles of N₂ will require = X moles of H₂
Solving for X,
X = (4 mol × 3 mol) ÷ 1 mol
X = 12 mol
Hence, 4 moles of N₂ requires 12 moles of H₂, but we are provided by 6 moles of H₂. Therefore, H₂ is a limiting reagent and will be consumed first.
<span>A) Nitrogen is in excess.
B) Hydrogen is in excess.
C) This is a synthesis reaction.
D) Hydrogen is the limiting reactant.
Answer:
Option-D (Hydrogen is the limiting reactant).
Explanation:
The equation of reaction is as follow,
N</span>₂ + 3 H₂ → 3 NH₂
According to this equation 1 mole of nitrogen reacts with 3 moles of hydrogen gas to produce 3 moles of ammonia.
In given illustration 4 moles of N₂ (blue spheres) are present and 6 moles of H₂ are present.
As,
1 mole of N₂ requires = 3 moles of H₂
So,
4 moles of N₂ will require = X moles of H₂
Solving for X,
X = (4 mol × 3 mol) ÷ 1 mol
X = 12 mol
Hence, 4 moles of N₂ requires 12 moles of H₂, but we are provided by 6 moles of H₂. Therefore, H₂ is a limiting reagent and will be consumed first.
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Answer:
.
Explanation:
<h3>Step One: calculate the coefficients. </h3>Look up the relative atomic mass of these four elements on a modern periodic table:
: approximately
.
: approximately
.
: approximately
.
: approximately
.
The relative atomic mass of an element is numerically equal to the mass (in grams, ,) of one mole of atoms of this element.
For example, the relative atomic mass of is approximately
. Therefore, each mole of
atoms would have a mass of
.
This sample contains of carbon. That would correspond to approximately
of
atoms.
Similarly, for the other three elements:
.
.
.
Hence, the ratio between these elements in this compound would be:
.
In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.
is indeed the smallest possible integer ratio between the number of atoms of these elements.
Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:
If carbon, , is present in this compound, then:
- The other elements in this compound will be listed in alphabetical order.
If there is no carbon in this compound, then list all the elements in this compound in alphabetical order.
Both (carbon) and
(hydrogen) are found in this compound. Therefore, the first element in the list would be
. The second would be
, followed by
and then
.
Hence, the empirical formula of this compound would be .
Answer:
Explanation:
The first step is:
Second step is:
Multiplying second step by 2, and adding both the steps, we get that:
Cancelling common species, we get that: