Clathrate Gun - 1

Chemistry Level pending

On the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or of raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. It is believed that if a sufficient amount of methane is released into the atmosphere, then the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.

Upon decomposition of $1.00\text{ g}$ of a methane hydrate with a fixed composition at $25^\circ C$ and atmospheric $(101.3\text{ kPa})$ pressure, $205\text{ mL}$ of methane is released.

Determine $n$ (not necessarily integer) in the formula of methane hydrate, $CH_4\cdot nH_2O$ to the nearest integer .

The answer is 6.

1 solution

Christopher Boo
Aug 5, 2014

By Ideal Gas Law , the amount of methane is given

$PV=nRT$

$(101300)(2.05\times10^{-4})=n(8.314)(25+273)$

$\boxed{n_{CH_4} = 8.382\times10^{-3}\text{ mol}}$

Now, we have to calculate the amount of $n_{H_2O}$ .

$m_{CH_4}=8.382\times10^{-3} \times 16=0.134g$

Thus,

$m_{H_2O} = 0.866g$

$\boxed{n_{H_2O}=4.811\times10^{-2}\text{ mol}}$

The ratio of amount of water and methane is

$\frac{4.811\times10^{-2}}{8.382\times10^{-3}} \approx 5.75$

Hence, the nearest integer is $6$ .

Clathrate Gun - 1

The answer is 6.

1 solution

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