The molecular formula of quinine is C20H<span>24N2</span>O<span>2. For every 1 mole of quinine molecule, there are 20 moles of carbon. Simply multiplying 6.0 moles by 20, we get, 120 moles.
Therefore, there are 120 moles of carbon in 6.0 moles of quinine.</span>
Answer:
Option D. 5.5
Explanation:
The equation is this:
2A + 6B ⇒ 3C
With the amounts that we were given, let's determine which is the <em>limting reactant</em>
2 A reacts with 6 B
4 A will react with ( 4 .6)/2 = 12B
I have 11 B, so the limiting is B
6 B react with 2 A
11 B will react with (11 .2 )/6 =3.66 A
I have 4 A, so A is the excess.
6 B produce 3 C
11 B will produce ( 11 .3)/6 = 5.5C
Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
Answer : The total change in enthalpy of this reaction is 25 kJ.
Explanation :
Enthalpy of reaction : It is defined as the changes in heat energy takes place when reactants go to products. It is denotes as .
ΔH = Energy of product - Energy of reactant
ΔH is positive when heat is absorbed and the reaction is endothermic.
ΔH is negative when heat is released and the reaction is exothermic.
In the given potential energy diagram, the energy of product at higher level and energy of reactant at lower level. The ΔH for this reaction will be positive.
Given:
Energy of product = 55 kJ
Energy of reactant = 30 kJ
ΔH = Energy of product - Energy of reactant
ΔH = 55 kJ - 30 kJ
ΔH = 25 kJ
Thus, the total change in enthalpy of this reaction is 25 kJ.
