Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
This is a missing part of your question:
The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given.
So you need the equilibrium balanced equation of SO2, O2, SO3 reaction:
First, we will start with the original equation which is not balanced yet (to understand how we get it):
SO2 + O2 ↔ SO3
Here the number of O atom is not equal at the to sides
So we will start to balance our equation by make the number of O atom equal each other on both sides:
So we will start to put 2SO3 instead of SO3
and put 2SO2 instead of SO2 to balance also the S atom on both sides
So we will get this:
2SO2(g) + O2(g) ↔ 2SO3(g) (This is our equilibrium balanced equation)
know we have a number of O atom equals on each side = 6
and the sulfur equals on each side = 2
M/V=D
16.52/2.26=D
Density=6.86 g/cm^3
Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.
