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### AuthorTopic: Hydrogen mass and percent yield-- I don't even know where to begin...  (Read 5905 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### pandabear

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##### Hydrogen mass and percent yield-- I don't even know where to begin...
« on: February 16, 2009, 03:19:51 PM »

Hydrogen can be extracted from natural gas according to the following reaction:
CH4(g) + CO2 (g)  ::equil::2 CO (g) +2 H2 (g)
Kp= 450 at 825  K
An 85.0-L reaction container initially contains 22.3 kg of CH4 and 55.4 kg of CO2 at 825 K.

Assuming ideal gas behavior, calculate the mass of  H2 (in g) present in the reaction mixture at equilibrium.

What is the percent yield of the reaction under these conditions?

There's so much jammed into one problem...can someone send me in the right direction?
thank you!
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#### Borek

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##### Re: Hydrogen mass and percent yield-- I don't even know where to begin...
« Reply #1 on: February 16, 2009, 09:45:51 PM »

Start from the beginning - calculate amount of hydrogen produced.

As you are given Kp convert known masses of reactants to their pressures in the reaction vessel.
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