Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Aqua Auras belong to the Quartz family. It's a synthetic type of Quarts that was enhanced with the help of gold coating, and sometimes other materials. The reason why it's called as aqua is because of its distinctive blue color thanks to the procedure which made it.
Your answer is sublimation.
Answer:
B
Explanation:
it shows that the solution is neutral.
I hope this helps
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>
