Answer:
0.252 mol
Explanation:
<em>Given the following reaction: </em>
<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>
<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>
First, we write the balanced equation.
Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂
We can establish the following relations.
- The molar mass of Cu is 63.55 g/mol.
- The molar ratio of Cu to Ag is 1:1.
The moles of Ag produced from 16.0 g of Cu are:
The reaction between ammonium sulfate and calcium hydroxide is given below.
(NH₄)₂SO₄ + Ca(OH)₂ --> 2NH₃ + CaSO₄ + 2H₂O
From the balance equation, we can conclude that every 74 g of calcium sulfate reacted with enough amount of ammonium sulfate will yield 34 grams of ammonia. From the given amount,
(20 g calcium sulfate) x (34 grams ammonia / 74 g calcium sulfate)
= <em>9.19 g ammonia</em>
Explanation:
When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.
Hence, ionic equation for this reaction is as follows.
Therefore, net ionic equation for this reaction is as follows.
Answer:
67.8%
Explanation:
La reacción de descomposición del CaCO₃ es:
CaCO₃ → CO₂ + CaO
<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>
Usando la ley general de los gases, las moles de dioxido de carbono son:
PV = nRT.
<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:
PV / RT = n
1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles
Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.
La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:
0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>
Así, la pureza del marmol es:
(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>
<h3>67.8%</h3>
