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2 years ago

In a synthesis reaction, 2.43g of aluminum is reacted with 9.22g of chlorine. Show your work using a BCA table.

Chemistry

1 answer:

Zinaida [17]2 years ago

6 0

Answer:

Chlorine gas is the limiting reactant

11.57 g

Explanation:

The equation of the reaction is;

2Al(s) + 3Cl2(g) -----> 2AlCl3(s)

The limiting reactant will produce the least number of moles of AlCl3.

For Al

Number of moles reacted = 2.43 g/27 g/mol = 0.09 moles

2 moles of Al yields 2 moles of AlCl3

0.09 moles of Al yields 0.09 * 2/2 = 0.09 moles of AlCl3

For Cl2

9.22g/71g/mol = 0.13 moles

3 moles of Cl2 yields 2 moles of AlCl3

0.13 moles of Cl2 yields 0.13 * 2/3 = 0.087 moles of AlCl3

Hence Cl2 is the limiting reactant

b) Mass of product produced = 0.087 moles of AlCl3 * 133g/mol = 11.57 g

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So the equation is going to be:
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11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

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By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

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Daniel [21]

Answer:

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