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Rom4ik [11]
3 years ago
12

Calculate the mass of 488 moles of calcium carbonate

Chemistry
1 answer:
san4es73 [151]3 years ago
6 0

Answer: 48800g

Explanation:

Using the mathematical relation : Moles = Mass / Molar Mass

Moles = 488

Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100g/mol

Therefore

488 = mass / 100 = 48800g

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En la reacción I2(g) + Br2(g) « 2 IBr(g), Keq = 280 a 150°C. Suponga que se permite que 0.500 mol de IBr en un matraz de 1.00 L
love history [14]

Answer:

P IBr: 15.454atm

I₂: 0.923 atm

P Br₂: 0.923atm

Explanation:

Basados en la reacción:

I₂(g) + Br₂(g) ⇄ 2 IBr(g)

La constante de equilibrio, Keq, es definida como:

Keq = \frac{P_{IBr}^2}{P_{I_2}P_{Br_2}}

<em>Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio</em>

Usando PV = nRT, la presión inicial de IBr es:

P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = <em>17.3 atm</em>

<em />

Siendo las presiones en equilibrio:

P IBr: 17.3 - 2X

P I₂: X

P Br₂: X

<em>Donde X representa el avance de reacción.</em>

Remplazando en Keq:

280 = (17.3 - 2X)² / X²

280X² = 4X² - 69.2X + 299.29

0 = -276X² - 69.2X + 299.29

<em>Resolviendo para X:</em>

X = -1.174 → Solución falsa. No existen presiones negativas

X = 0.923 → Solución real

Así, las presiones parciales en equilibrio de cada compuesto son:

P IBr: 17.3 - 2X = <em>15.454atm</em>

P I₂: X =<em> 0.923atm</em>

P Br₂: X = <em>0.923atm</em>

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