How many grams of KO2 are needed to form 6.5 g of O2?
Chemistry · High School · Thu Feb 04 2021
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To find out how many grams of KO2 (potassium superoxide) are needed to produce 6.5 g of O2 (oxygen), we need to first write down the chemical reaction and then use stoichiometry to do the calculation.
The reaction for the decomposition of KO2 is: 4 KO2 → 2 K2O + 3 O2
To calculate the amount of KO2 needed to produce 6.5 g of O2, we need to know the molar mass of O2 and KO2. The molar mass (MM) of O2 is 32 g/mol (since O has an atomic mass of 16 g/mol and O2 has two O atoms). The molar mass of KO2 is approximately 71 g/mol (K has an atomic mass of 39 g/mol, O has an atomic mass of 16 g/mol, and KO2 has one K atom and two O atoms).
Steps: 1. Calculate the moles of O2: grams of O2 / MM of O2 = moles of O2 6.5 g / 32 g/mol = 0.203125 mol O2
2. Use the molar ratio from the balanced chemical equation to find moles of KO2: From the reaction, we know that 3 moles of O2 are produced from 4 moles of KO2. So, the molar ratio is 4 moles KO2 : 3 moles O2.
3. Calculate the moles of KO2: (moles O2) × (4 moles KO2 / 3 moles O2) = moles KO2 0.203125 mol O2 × (4/3) = 0.270833 mol KO2
4. Calculate the grams of KO2: moles of KO2 × MM of KO2 = grams of KO2 0.270833 mol KO2 × 71 g/mol = 19.229145 g KO2
So, you would need about 19.23 grams of KO2 to form 6.5 g of O2.