Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
Answer:The product and reactants reach a final, unchanging level.
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Answer:
Explanation:
Hello,
In this case, the reaction is given as:
Thus, starting by the yielded grams of silver iodide, we obtain:
Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:
And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.
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Answer:
The theoretical yield of
Li
3
N
is
20.9 g
.
Explanation:
Balanced Equation
6Li(s)
+
N
2
(
g
)
→
2Li
3
N(s)
In order to determine the theoretical yield, we must first find the limiting reactant (reagent), which will determine the greatest possible amount of product that can be produced.
Molar Masses
Li
:
6.941 g/mol
N
2
:
(
2
×
14.007
g/mol
)
=
28.014 g/mol
Li
3
N
:
(
3
×
6.941
g/mol Li
)
+
(
1
×
14.007
g/mol N
)
=
34.83 g/mol Li
3
N
Limiting Reactant
Divide the mass of each reactant by its molar mass, then multiply times the mole ratio from the balanced equation with the product on top and the reactant on bottom, then multiply times the molar mass of
Li
3
N
.
Lithium
12.5
g Li
×
1
mol Li
6.941
g Li
×
2
mol Li
3
N
6
mol Li
×
34.83
g Li
3
N
1
mol Li
3
N
=
20.9 g Li
3
N
Nitrogen Gas
34.1
g N
2
×
1
mol N
2
28.014
g N
2
×
2
mol Li
3
N
1
mol N
2
×
34.83
g Li
3
N
1
mol Li
3
N
=
84.8 g Li
3
N
Lithium produces less lithium nitride than nitrogen gas. Therefore, the limiting reactant is lithium, and the theoretical yield of lithium nitride is
20.9 g
.
Explanation:
