Given 4K(s) + O2(g) ---> 2K2O(s), how many grams of K2O(s) can be produced if 3.91 grams of K(s) are allowed to react with 1.60 grams of O2(g)?
The equation of the chemical reaction between potassium (K) and oxygen (O2) is: 4K + O2 --> 2K2O
From the equation it is seen that 4 moles of potassium react with with one mole of oxygen gas (O2) to give 2 moles of potassium oxide.
The molar mass of potassium is 39.09 g/mole. 3.91 g of potassium is equivalent to approximately 0.1 moles of potassium. The molar mass of oxygen is 32 and 1.6 g of O2 is equivalent to 0.05 mole. As 4 moles of potassium are required to react with 1 mole of oxygen, potassium is the limiting reagent here. The number of grams of K2O formed is equal to 4.71 g.
When the reaction is complete, all the potassium K(s) is used to form K2O (s). 0.025 moles of oxygen gas O2(g) is present in the reactant side as there is not enough K(s) for it to react.