A 32.8 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?

Chemistry

1 answer:

olasank [31]3 years ago

8 0

Answer:

32.8g/mole

Explanation:

Given parameters:

Mass of sample of gas = 32.8g

Volume = 22.4L

Unknown:

Molecular weight = ?

Solution:

To solve this problem we must understand that at rtp;

1 mole of gas occupies a volume of 22.4L

Number of mole of the gas = 1 mole

Now;

Mass = number of moles x molecular weight

molecular weight = $\frac{mass}{molecular weight}$ = $\frac{32.8g}{1mole }$ = 32.8g/mole

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The question is incomplete, here is the complete question:

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Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Answer: The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

Explanation:

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$