Answer:
0.145 moles de AlBr3.
Explanation:
¡Hola!
En este caso, al considerar la reacción química dada:
Al(s)+Br2(l)⟶AlBr3(s)
Es claro que primero debemos balancearla como se muestra a continuación:
2Al(s)+3Br2(l)⟶2AlBr3(s)
Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:
Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.
¡Saludos!
Answer:
ksp = 0,176
Explanation:
The borax (Na₂borate) in water is in equilibrium, thus:
Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)
<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>
The ksp is defined as:
<em>ksp = [borate²⁻] [Na⁺]²</em>
Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:
B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻
The moles of HCl that reacts with B₄O₇²⁻ are:
0,500M×0,01200L = 6,00x10⁻³ mol of HCl
As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:
6,00x10⁻³ mol of HCl× = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>
For (1), moles of Na⁺ are <em>3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺</em>
The [borate²⁻] is <em>3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = </em><em>0,353M</em>
And [Na⁺] is <em>6,00x10⁻³ mol of Na⁺ / 0,00850L = </em>0,706M
Replacing in the expression of ksp:
ksp = [0,353] [0,706]²
<em>ksp = 0,176</em>
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I hope it helps!