Question

Which is the lim iting reactant in the following reaction given that you start with 42.0 of CO2 and 99.0 g of KOH? Calculate the percent yield if 115 g of K2CO; is actually recovered. 4 pts Reaction: CO2 + 2KOH K2 CO3 + H20

Answer 1

: 1 O2 + 2KOH -- Ke C03 + H2O No. of moles of CO₂ = Mass = 42.09 . Molar Mass 44.olglimal = 0.954 mal No of moles of kot= 99.0g = 1.76 mal... - 56.10 5691mof Now 1 mole of CO2 reacts completely with 2 males of Kor. .. 0.954 mole of (0g would require 0.954 X2=1.908 mole Kol. But as no. of moles of Kot available are less than this . kah a the limiting reagent and thus Koh will determine the o amount of product that can be formed Also 2 moles of Koh gives 1 mole Karuz o No of males of Ke (Oz that can be formed = 1.76 = 0.880 mole Theoretical Yield of ReCO3 = No of moles x Molar mass = 0.880 mof x 138:2059/mat - = 121.69 Also; % lielda Obtained Yield x 100% Theoretical Yield = 1154 x 100.. = 94. 6.10 t L 121.09