68% 11/07/17 0 This is a Numeric Entry question/ It is worth 4 points/You have 4

Question

68% 11/07/17 0 This is a Numeric Entry question/ It is worth 4 points/You have 4 of 5 attempts remaining/There is no attempt penalty 02 Question (4 points) Q See page 302 Potassium superaxide KOa reacts with carbon dioxide to form potassium carbonate and oxygens 4K02 + 2002 2K2co,+ 302 2nd attempt hi See Periodic Table This reaction makes potassium superoxide useful in a self contained breathing apparatus. How much O, could be produced from 2.578 of KO2 and 4.46 g of CO, > 1st attempt OF 15 QUESTIONS COMPLETED K02/15 3 4 5 6 8 9

Explanation / Answer

Balanced equation:
4 KO2 + 2 CO2 = 2 K2CO3 + 3 O2
Reaction type: double replacement

2.57 gm of KO2 = 2.57 / 71.09 =  0.03614 Moles

4.46 gm of CO2 = 4.46 / 44 =  0.10111 Moles

Limiting reagent is KO2

Moles of O2 to be produced =  0.02711 Moles

Mass of O2 to be produced = 0.02711 x 32 =  0.8675 gm

Question 2

Balanced equation:
6 HF + 3 NaAlO2 = Na3AlF6 + 3 H2O + Al2O3

3.91 Kg of Na3AlF6 = 3910 / 209.94 =  18.62 Moles

Moles of NaAlO2 needed =  55.87 Moles

Mass of NaAlO2 needed =  55.87 x 81.97 = 4579.89 gm or 4.579 gm .

Question 4

Mass of carbon = 6.55

Moles of carbon = 6.55 / 12 =  0.5453 Moles

Moles of CO2 to be produced =  0.5453 Moles

Mass of CO2 =  0.5453 x 44 = 24 gm

Percentage yield = 14.4 x 100 /24 = 60 %

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