Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of butane
b) moles of oxygen
According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
Answer:
pH = 12.22
Explanation:
<em>... To make up 170mL of solution... The temperature is 25°C...</em>
<em />
The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:
Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)
<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>
<em />
To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:
<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>
0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =
2.80x10⁻³ moles of OH⁻
<em>Molarity [OH⁻] and [H⁺]</em>
2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M
As Kw at 25°C is 1x10⁻¹⁴:
Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]
[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M
<em>pH:</em>
pH = -log [H⁺]
pH = -log [6.068x10⁻¹³M]
<h3>pH = 12.22</h3>
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>
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
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