Question

We are supposed to show our work using conversion factors but no youtube videos are really helping me to be able to figure this one out, can anyone help?

12. Potassium superoxide, KO. can be used as an emergency source of oxygen in space vehicles because of the reaction below. 4 KO2 + 2 CO2 → 2 K2CO3 + 3 Oz a. (8 pt) If 25.00 g KO, (MM = 71.10 g/mol) reacts with 12.50 g CO: (MM = 44.01 g/mol). what is the theoretical yield, in grams, of Oz (MM = 32.00 g/mol)? b. (6 pt) How many grams of each reactant will remain when the reaction is complete? a ko Page 6 of 9

Answer 1

Huko +200 -72 kcat 302_1 . moly 25 12.5ou: 71.10. 44 = 0.3516 2840 molest mole of ko react with 1 mole 2 of a so moles of co reinere for 0.3576 тър 40, Ix0.3516 = 0.057 moley Thence here ka is limiting reagent each ko, mole podne 3 mole of a I so 0.3516 mole of bo produce 3 x 0.3516 of 2 i 0.2637 mole ore Scanned with CamScanner Theoretical yield in gon) of a a mole x molar bay P a 0.2832x32 TS 804384 g of o amount of reautant left ko. is completely consumed but co is lo left Lamoat of co, left = (0.2840 -0.128) mok = 0.1882 moles in goma 0.1002844.01 - 4. 168gm ca left cscanned with CamScanner