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mafiozo [28]
3 years ago
15

A chemistry student needs 90.00 g of thiophene for an experiment. She has available 0.50 kg of a 12.3% w/w solution of thiophene

in ethanol.
Calculate the mass of solution the student should use. If there's not enough solution, press the 'No solution" button
Round your answer to 3 significant digits. Put the answer in terms of the amount of grams.
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
7 0

<u>Answer:</u> The correct answer is No solution.

<u>Explanation:</u>

We are given:

Given mass of thiophene for the experiment = 90.00 g

12.3% w/w solution of thiophene

This means that 12.3 g of thiophene is present in 100 g of solution

Applying unitary method:

If 12.3 g of thiophene is present in 100 g of solution

So, 90.00 g of thiophene will be present in \frac{100g}{12.3g}\times 90.00g=731.7g of solution

Converting it into kilograms:

1 kg = 1000 g

So, 731.7 g = 0.7317 kg

As the given amount of solution is 0.50 kg which is less than the required amount.

Thus, there is not enough solution the student should use.

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H₂O = 3.56 moles

MgCl₂ = 1.78 moles

Explanation:

7 0
3 years ago
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
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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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3 years ago
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