I need help on this question. Could someone give me tips on how to solve it?
thanks
An emergency breathing apparatus often placed in mines works via the following chemical reaction: 4KO2(s) + 2CO2(g) to make 2K2CO3(s) + O2(g)
If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breath while exiting the mine. Notice that the reaction produces O2, with can be breathed, and absorbs CO2, a product of respiration. What minimum amount of KO2 is required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit in an emergency? Assume that an adult consumes approximately 4.4g of oxygen in 15 minutes of normal breathing.
4.4g of O2? figure the moles of O2, then
Look at the balanced equation: it takes four times that amount of moles of O2 of KO2. What is the mass of that?
To solve this problem, you need to follow these steps:
1. Calculate the number of moles of O2 required for an adult to breathe in 15 minutes.
- Given that an adult consumes approximately 4.4g of O2, you can convert this mass to moles using the molar mass of O2 (32 g/mol).
- Moles of O2 = Mass of O2 (g) / Molar mass of O2 (g/mol)
2. Use the balanced chemical equation to determine the stoichiometric ratio between O2 and KO2.
- The balanced equation tells us that 4 moles of KO2 produce 2 moles of K2CO3 and 1 mole of O2.
- From this ratio, we can conclude that 4 moles of KO2 are needed to produce 1 mole of O2.
3. Convert the moles of O2 to moles of KO2.
- Since we now know the stoichiometric ratio, we can multiply the moles of O2 by the ratio to obtain the moles of KO2.
- Moles of KO2 = Moles of O2 x (4 moles of KO2 / 1 mole of O2)
4. Convert the moles of KO2 to the mass of KO2.
- To obtain the mass of KO2, multiply the moles of KO2 by the molar mass of KO2.
- Mass of KO2 = Moles of KO2 x Molar mass of KO2 (g/mol)
By following these steps, you should be able to determine the minimum amount of KO2 required for the emergency breathing apparatus.