##### A mixture of helium and carbon dioxide gases is maintained in a 6.63 L flask at

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A mixture of helium and carbon dioxide gases is maintained in a 6.63 L flask at a pressure of 3.68 atm and a temperature of 83 °C. If the gas mixture contains 0.756grams of helium, the number of grams of carbon dioxide in the mixture is  g.

Jun 1st, 2015

the number of grams in the mixture is 6.63

pressure at 3.68 atm

temperatures 83degrees

6.63/3.68*0.756=1.36 grams

Jun 1st, 2015

wrong

Jun 1st, 2015

not really you are making mistakes yourself this is the right way.

Jun 1st, 2015

no see.

The number of moles of He can be calculated from the number of grams of He and the molar mass:

 0.756 g He [img src="http://cxp.cengage.com/contentservice/assets/T=1432666732132/owms01h/mediaarchives/GenChem/Image/times14-3.gif"> =  0.1889 mol He 4.003 g

The number of moles of CO2 can then be calculated from the total number of moles:

nCO2 = ntotal - nHe = 0.8352 - 0.1889 = 0.6463 mol CO2

And finally, the number of grams of CO2 can be calculated:

 0.6463 mol CO2 [img src="http://cxp.cengage.com/contentservice/assets/T=1432666732132/owms01h/mediaarchives/GenChem/Image/times14-3.gif"> 44.01 g =  28.4 g CO2 mol

Jun 1st, 2015

ok i will correct my source then, i thought that the way its done.

Jun 1st, 2015

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Jun 1st, 2015
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Jun 1st, 2015
Dec 3rd, 2016
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