How many grams of o2 are needed in a reaction that produces 75 g of na2o?

Suppose you're talking about the following reaction: 4 Na + O2 ---> 2 Na2O If this is not the reaction then just keep reading, you'll find out how to calculate it your self! This is an pretty easy question because it involves a simple trick. With this reaction you're goal is to have 75 gram's Na2O. If the reaction, the molar weight's, one amount in grams of one product or reactant beside oxygen, and the stoichometry is know we can calculate the grams of oxygen. The molar weight of Na2O is 61,97893856 g/mol O2 is 31,9988 g/mol we have 75 gram Na2O divided by 61,97793856 it brings us 1,21 mol. Because the S.U unit mole is a variabel for the amount of molecules we can see that, because 2 Na2O molecules form from 1 molecule O2 and four molecules Na, that dividing 1,21 by 2 brings us the amount of molecules in mole for O2. Keep in mind that the mole is just a value that stands for an amount of molecules This is because an amount of molecules in one mole weigh 31,9988 gram. That is the molair weight!!! So for our final conclusion I present to you THE ANSWER :) (1,21/ 2) * 32 = around 19,4 grams!

50 grams of O2 is needed because for every one part of Na, two parts of O2 are needed. You can just take 75 and divide it by three then multiply by 2

13=7.8514=3.5feet15=10.99feet16=2& 317=418=about 9ft19=7.8520=4in

step-by-step explanation: