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Mercury can be obtained by reacting mercury(II) sulfide with calcium oxide. How many grams of calcium oxide are needed to produce 36.0 g of Hg?4HgS(s) + 4CaO(s) ® 4Hg(l) + 3CaS(s) + CaSO4Group of answer choices13.4g7.56g10.1g1.80g

Question

Mercury can be obtained by reacting mercury(II) sulfide with calcium oxide. How many grams of calcium oxide are needed to produce 36.0 g of Hg?4HgS(s) + 4CaO(s) ® 4Hg(l) + 3CaS(s) + CaSO4Group of answer choices13.4g7.56g10.1g1.80g

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Solution

Para resolver este problema, sigamos los siguientes pasos:

  1. Escribir la ecuación balanceada: 4HgS(s)+4CaO(s)4Hg(l)+3CaS(s)+CaSO4(s) 4 \text{HgS} (s) + 4 \text{CaO} (s) \rightarrow 4 \text{Hg} (l) + 3 \text{CaS} (s) + \text{CaSO}_4 (s)

  2. Determinar la masa molar de cada compuesto involucrado:

    • Hg (mercurio): 200.59 g/mol
    • CaO (óxido de calcio): 40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol
  3. Calcular los moles de Hg producidos: Moles de Hg=36.0 g200.59 g/mol0.1795 moles \text{Moles de Hg} = \frac{36.0 \text{ g}}{200.59 \text{ g/mol}} \approx 0.1795 \text{ moles}

  4. Usar la estequiometría de la reacción para encontrar los moles de CaO necesarios: La ecuación balanceada muestra que la relación molar entre Hg y CaO es 1:1. Por lo tanto, se necesitan 0.1795 moles de CaO para producir 0.1795 moles de Hg.

  5. Calcular la masa de CaO necesaria: Masa de CaO=Moles de CaO×Masa molar de CaO \text{Masa de CaO} = \text{Moles de CaO} \times \text{Masa molar de CaO} Masa de CaO=0.1795 moles×56.08 g/mol10.1 g \text{Masa de CaO} = 0.1795 \text{ moles} \times 56.08 \text{ g/mol} \approx 10.1 \text{ g}

Por lo tanto, se necesitan aproximadamente 10.1 g de CaO para producir 36.0 g de Hg. La respuesta correcta es 10.1g.

This problem has been solved

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