Question

please help with the calculations and limiting reagent\percent yeild!!!

EXPERIMENT REPORT SHEET Synthesis of KzFe(C204)3.3 H20 15.0 mL 4a5a watch glass 43.958 : paper Experimental Data Volume of FeCl, solution Mass of K C,04H2O Mass of product 47.891 Calculations Moles of FeCl3 Moles of K C2O4 H2O Moles of product (KsFe(C204)3.3 H20) Limiting reagent Percent yield Comment on possible impurities in your product: Show all work for determining the limiting reagent and percent yield.

molarity of FeCl3 is .40 g/mol

we had 5.0 mL of FeCl3 (0.40g/mL) which equals 0.08g divided by 162.2g FeCl3 which equals 0.000493 moles or 4.9 x 10^-4

Answer 1

Chemical equation of preparation, 3 kg COy. It go + fellz (kq) - 184.2291mol 162.2 g/mol kz fel CaOu)z. 3 H₂O (49) + 3 kcl (aq) 491.21g/mol Volume of felly used = 5ml mass of felly = volume x density = S(x 0.90g = 2 g fella ME Mass(g) moles of fella = 29 molar mass - 162.29 Imol. = 0.0123 moles. (glmol) Moles 8.00 coles. mass of Kg CgOy. Hgo used : 8.00g moles of Kg 24 H2O = sig dog line = 0.04343 moles. for 0.0123 moles of felly we require 3 mol-Ca Kg Oy . HgO x 0.0123 mot feltz I mot feltz = 0.0309 moles of K₂ C₂O4. Hao will need. and we have already 0.04343 moles of kg Oy to available, thus fellz is the limiting reagent (decides the mass of product). 0.0123 moles of felly will form = 0.0/23 moles of product Kzfe (204); 3430 theoretical Mam of product = moles x molas maks = 0.0123 motes x 491.219., Actual yield of product = 47.891-43.958 = 3.9339 d. Actualyield 4100 3.9339 100 - 6514 theoretica lyield 6.0429 This t. yield : 65.1%. u91.219=6.0429 mot