A cylinder of volume 1m3 is filled with 1kg of CO and 100g of H2O. What is the molar percentage of CO in the mixture in the cylinder?

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A cylinder of volume 1m3 is filled with 1kg of CO and 100g of H2O. What is the molar percentage of CO in the mixture in the cylinder?

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Answer 1 · 2018-04-10T06:07:15

87%

Explanation:

in 1 kg of CO you have $n = \frac{g}{M M} = \frac{1000 g}{28 \frac{g}{m o l}} = 35, 7 m o l$ $n= g/(MM)=(1000g)/(28g/(mol))=35,7 mol$
in 100 g of $H_{2} O$ $H_2O$ you have $n = \frac{g}{M M} = \frac{100 g}{18 \frac{g}{m o l}} = 5, 5 m o l$ $n= g/(MM)=(100g)/(18g/(mol))=5,5 mol$
total mol = 35,7 + 5,5 = 41,2 mol
Molar percentage of CO = $\frac{m o l C O}{m o l_{t}} \times 100 = \frac{35.7}{41.2} = 87 %$ $(mol CO)/(mol_(t)) xx 100 =35.7/41.2=87%$

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A cylinder of volume 1m3 is filled with 1kg of CO and 100g of H2O. What is the molar percentage of CO in the mixture in the cylinder?

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