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andriy [413]
3 years ago
10

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

lcio que se obtiene al hacer reaccionar cantidades iguales de calcio y oxígeno (por ejemplo, 50 g de cada uno)
Chemistry
2 answers:
steposvetlana [31]3 years ago
7 0

Answer:

69.9 g of CaO will be produced. ≅ 70 g

Explanation:

First of all you need to make the reaction:

2Ca + O₂ → 2 CaO

Determine the moles of each reactant:

10 g Ca / 40.08 g/mol = 0.25 moles

4 g O₂ / 32 g/mol = 0.125 moles

There is no limiting reagent in this reaction, we can use both elements.

Ratio between Ca and CaO, is 2:2. For 0.25 moles of Ca I would make 0.25 moles of CaO. We convert the moles to mass:

0.25 mol . 56.08 g / 1mol = 14 g

Let's think when you have the same mass of reactant:

50 g Ca/ 40.08 g/mol = 1.24 moles

50 g O₂ / 32 g/mol = 1.56 moles

For 1 mol of oxygen I need 2 moles of calcium, so If I have 1.56 moles, I would need the double, 3.12. Notice that Ca is the limiting reagent (we need 3.12 moles of reactant, but we only have 1.24). Now we need to work with it. If 2 moles of Ca, makes 2 moles of CaO, then 1.24 moles, will produce the same amount of oxide. We finally convert the moles to mass: 1.24 mol . 56.08 g/mol = 69.9 g

egoroff_w [7]3 years ago
4 0

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

2Ca+O_2\rightarrow 2CaO

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO

Esto quiere decir que de 50 gramos de oxígeno, solo 20 gramos reaccionan para formar 70 gramos de óxido de calcio.

Saludos!

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Answer:

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Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

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The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

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The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

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In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

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