Question

what is the percent yield a precipitate formed when 15.4 mL of 1.25 molar copper two nitrate is mixed with 25.3 mL of 0.975 molar sodium hydroxide if the theoretical yield is 0.752 g

Answer 1

Given in Volume of Cul NO₂ = 15.4 mL or 0.0154L. molarity of Cultos) = 1,25 m Hence number of moles of Cu(N/2= na molasiti x volume n= 1.25% 0.0154 n= 0.01925 moles. volume of NaOH = 25.3 mL or 0.0253 L molarity of NaOH = 0.975 m Hence number of moles of Nao T na molarityx volume no 0.975 X 0.0253 n=0.02466 moles. Rxn involved: CULNO3)2 + 2 NOWH CucoH) + 2 Nano, 0.01925(moles) 0.024661 milu) Here NaOH would be limiting reagent because reaction is depend on it hence reaction would be CuCNO3)₂ + 2NaOH to culor) + 2Nanoz. The reaction in 61:2:1) means I part & 2 part of CHINO3)2 & Naoh reachs & I part of a colle formed

Cu(10), + 2 MOOH - ) Cu(OH), t 2 MANO3 0.02466 Hven 0.01425 'melh 0.01233 mole 0.02464 Quluneet 0.01233 0.0246 2.024.66 . o moles Lemin moly 0.01233 moles 0.01425 -0.012 33 0.00192 metodo molecular weight of Cu(OH) = 97.57g/mol. Hence 0.01233 moly of Culon), produced » 0.01233(moles)*97.573 /molens = 1.2030 g % yield • Actual yield. xiao Theostical yield Giren : Theortical yield= 0.7529 Actual yield = 1.2030 g % yields. Actualyield x100 Theofticafyield 1:20.30 x100 % yilld. = 159.91% 0.752