Question

1. (12 pts) For the balanced equation below, determine how many grams of UF4 (MW: 314.0 g/mol) can be produced when 7.31 kg of UO2 (MW: 270.0 g/mol) react with 3.94 L of 3.54 M HF (MW: 20.01 g/mol). UO2(g) + 4 HF(aq) → UF4(g) + 2 H20(1) 2. (12 pts) A) Determine how much chromium metal can be produced when 284.54 g of chromium (III) oxide reacts with excess aluminum according to the reaction below. Cr2O3(s) + 2 Al(n) = 2 Cr() + Al2O3(s) B) If 23.16 g of chromium metal results, what is the percent yield for the reaction?

Answer 1

Mole of vo = 7.31 x 10² 270 à 27 mole Mole of HF = 3.94 x 3.54 = 13.95 mole Here Hf is limiting reagent because its mole. is less coefficient ratio Mole of ufy formed = mde of HF Coefficient of HF 13.95 – 3.49 nde Mass of ufy formed = Mole x Modlar mass - 3.49 x 310 gm . 1094.89 gm - 1.094 kg

C%2O3 + 2A1 - 200 t. Al2O Molar mass of Cr2O3 = 152 gm Mole of (203 reacted = 284.54 152 = - 1.87 mole Mole of cr formed - Coefficient of co x Mole of crgo, 2 x 1.87mole = 3.74 mole (r metal produced = Mole of cox Molar Mass of cr = 3.74 x 52 = 194.48 gen Mas of chromium metal result mass of cometal expected x 100 23.16 100 - 194.48 - 11.9 %
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