To calculate the moles of O2 produced, first find the moles of CO2 using its molar mass, which is 44.01 g/mol. Then, use the mole ratio from the balanced equation to find the moles of O2 produced. Finally, multiply the moles of CO2 by the mole ratio to get the moles of O2 produced.
KO2 is rare and is called potassium superoxide. It consists of K+ ion and O^2- ion. In this case, the oxidation number of O would be 1-. If, however, you meant to ask about K2O (potassium oxide), then the oxidation number of O would be 2-
The chemical symbol for superoxide is O2-.
Potassium is the alkali metal that can react with oxygen to form peroxide (K2O2) or superoxide (KO2).
Potassium superoxide (KO2) is used in submarines to generate oxygen. When KO2 reacts with water vapor in the air, it releases oxygen gas. This provides a source of breathable air for the crew in case of emergencies or when the submarine is submerged for extended periods.
One mole of KO2 produces 1 mole of O2. The molar mass of KO2 is 71.1 g/mol and that of O2 is 32 g/mol. So, 8.5 g of O2 would require (8.5 g / 32 g/mol) x 71.1 g/mol = 18.95 grams of KO2.
For every 4 moles of KO2 consumed, 3 moles of O2 are produced. Since the molar mass of KO2 is 71.097 g/mol, 323.2 grams of KO2 is approximately 4.55 moles. Therefore, 3/4 * 4.55 = 3.41 moles of O2 are produced. This is equivalent to 6.83 grams of O2.
The chemical formula of potassium superoxide is KO2.
The balanced chemical equation for the reaction would be: 4KO2 + 2CO2 → 2K2CO3 + 3O2 Since the molar ratio between KO2 and O2 is 4:3, 0.400 mol of KO2 would produce: 0.400 mol KO2 * (3 mol O2 / 4 mol KO2) = 0.300 mol O2
To determine the amount of KO2 required to produce 1120.0 L of O2 at 20.0°C and 1.00 atm, we need to use the ideal gas law. First, calculate the moles of O2 using PV = nRT. Then, use the balanced chemical equation for the decomposition of KO2 to O2 to find the molar ratio and the mass of KO2 needed.
KO2
To calculate the moles of O2 produced, first find the moles of CO2 using its molar mass, which is 44.01 g/mol. Then, use the mole ratio from the balanced equation to find the moles of O2 produced. Finally, multiply the moles of CO2 by the mole ratio to get the moles of O2 produced.
The minimum amount of KO2 required for the apparatus to produce enough oxygen for a 15-minute emergency exit depends on the flow rate of oxygen required by the user and the reactivity of KO2. Typically, a 500g cartridge of KO2 can provide enough oxygen for 15-30 minutes of use in emergency situations. It's essential to test the apparatus to determine the exact amount needed for 15 minutes.
The balanced equation for KO2 + H2O is 2KO2 + 2H2O -> 2KOH + O2. This means that for every 2 moles of KO2 and H2O, you get 2 moles of KOH and 1 mole of O2.
The oxidation number for oxygen (O) in KO2 is -1. Since the overall charge of the compound is 0, the oxidation number of potassium (K) is +1.
When potassium and oxygen react, the most stable compound formed is potassium oxide (K2O). KO2 can be formed when there is an excess of oxygen present. In this case, oxygen binds to potassium to form potassium superoxide (KO2) instead of the normal oxide.
K + o2 = ko2