Considering the limiting reactant, what is the mass of iron produced from 80.0 g of iron(II)oxide (71.55 g/mol) and 20.0 g of ma
1 answer:
Answer:
45.95 grams of iron
Explanation:
The reaction between FeO and magnesium metal is
1. The first step is verify that the equation is balanced. In this case it is balanced since the numbers of atoms of each element are equal in both sides of equation (for example we have 1 Fe atom in reactants and 1 Fe atom in products)
2. We need to calculate the limiting reactant
- for that we need to calculate the number of moles of each reactant dividing the mass that we have by the respective molecular weight of the compound.
and
the molecular weight of Mg (24.305) can be readed in the periodic table of elements.
- so we divide the moles by stoichiometry number (number in front of each compound in the equation) in this case is 1 for both reactants (that is we need 1 mol of FeO and 1 mol of Mg to produce 1 mol of Fe).
- The lower number obtained was 0.823 for Mg, so Mg is the limiting reactant.
3. We calculate the moles and the mass of Fe produced. The maximum number of moles of Fe that can be produced is given by the limit reactant. So we would use the moles of Mg to calculate the Fe produced.
- We have 0.823 mol of Mg and the chemical equation shown above say that we need 1 mol of Mg to produce 1 mol of Fe, so with 0.823 mol of Mg we woud produce 0.823 mol of Fe (
).
- To convert from mol of Fe to grams of Fe we would multiply by the molecular weight of Fe
(molecular weight of Fe is readed in the periodic table of elements). So it is produced 45.95 g of iron
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The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸
Construct ICE table and obtain their equilibrium concentrations:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M): 0.1 0 0
C (M): -x +x +x
E (M): 0.1 -x x x
So:
9.0 x 10⁻⁸ =
x = 9.4 x 10⁻⁵
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M): 9.4 x 10⁻⁵ 0 9.4 x 10⁻⁵
C (M): -x +x +x
E (M): 9.4 x 10⁻⁵ -x x 9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ =
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸
Construct ICE table and obtain their equilibrium concentrations:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M): 0.1 0 0
C (M): -x +x +x
E (M): 0.1 -x x x
So:
9.0 x 10⁻⁸ =
x = 9.4 x 10⁻⁵
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M): 9.4 x 10⁻⁵ 0 9.4 x 10⁻⁵
C (M): -x +x +x
E (M): 9.4 x 10⁻⁵ -x x 9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ =
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M
Answer: The electronic configuration of potassium atom in excited state is
Explanation:
There are 2 states classified under energy levels:
1. Ground state: This is the lower energy state which is termed as stable state.
2. Excited state: This is the upper energy state and is termed as the unstable state. All the electrons which are present in this state always come back to the ground state.
Ground state electronic configuration of potassium (Z = 19) :
As, only 1 electron can freely move to upper state. Hence, the excited state electronic configuration of potassium will be:
