Describe how we can determine the molar volume of gas in Silver oxide decomposes

Question

Describe how we can determine the molar volume of gas in Silver oxide decomposes into silver metal and oxygen gas. A student collects the following data mass of crucible is 30.296g mass crucible/e plus silver oxide is 38.623g mass of crucible and contents after reaction has occurred is 38.061 g What the balanced equation? What is the starting mass of silver oxide? What is the mass of oxygen released? What is the experimental percentage of oxygen from the data? What is the theoretical percentage of oxygen in silver oxide from the formula? is the percent error?

Explanation / Answer

Question a

Balanced equation:
2 Ag2O(s) ===> 4 Ag(s) + O2(g)

Question b

Mass of silver oxide = 38.623 - 30.296 = 8.327 gm

Question C

8.327 gm of Ag2O = 8.327 / 231.73 = 0.03593 Moles

0.03593 Moles of Ag2O will produce 0.01796 moles of O2

Mass of oxygen = 0.01796 x 32 = 0.5749 gm

Mass of Ag obtained = 38.623 - 38.061 = 0.562 gm

Experimental Percentage of oxygen = 0.562 x 100 / 0.5749 = 97.75 %

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