Question

PART II: (50-Pts) Demonstrate Your Problem Solving Skills Taught/Learned in C Remember, you must show your work to earn any partial credit on any of the giv- 1) (6-pts) In the reaction below, potassium hydroxide and oxygen gas are formed 4KO2(s)+2H2O (I) (SHOW CALCULATIONS TO EARN ANY POINT) (8) O€(H a) If 60.00 g of KO2 is reacted with 15.00 g H20, what is the limiting reactant? Coo.004 71.0 48 9mal O.3439 ma k02 Orot X 07H many grams of KOH(s) will be produced? HOH

Answer 1

HKO2 + 2H₂O 4 KOH + 302 umoles of Ko, reals weth 2 moles of Heo 4(71) gm of Ko, reacts with 36gm of , 60 gms 1 u 2 60 x 36 = 7.60gm of Hbo required, 284 but given as 15gm of H₂o. limiting reageut = k02 umoles KO, → umol KOH Amount of ko, 4 x 71m Ko 56 gm kot | 60 qm 279 200 X56=11.832