We are given with the equation 2Na + O2 → 2Na2O where Na and O2 react to form sodium oxide. We are also given with two amounts 55.3 g Na with 64.3 g O2 .In this case, we will find the limiting reactant. we divide each with the molar mass and the stoich coeff. Na is 1.202, O2 is 2.01. Hence the limiting is Na. we base the calculations here.The amount is equal to 46.88 grams
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make. start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
<span>Ca + Cl2 = CaCl2 </span> We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.
56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction
