am 27n) 1 kg m /s? e Eler Problem 5. (10 points possible plus 5 bonus points): P

Question

am 27n) 1 kg m /s? e Eler Problem 5. (10 points possible plus 5 bonus points): Perform the calculation and write your final answers in the answer boxes provided in the form (units etc.) specified in the problem. Also, be sure to show all of your work. CaF, H2SO4 Caso t 2HF (Note: MM values are 78.07 g/mol for CaF2, 98.08 g/mol for H2SO4, 136.14 g/mol for Caso4, and 20.01 g/mol for HF.) Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction. In one process, 6.00 kg of CaF2 are treated with 7.30 kg of H2SO4 and yield 2.86 kg of HF (a) Calculate the theoretical yield of HF. (State your fin answer in kg with 3 sig figs.) no (b) Calculate the percent yield of HF. (State your final answer in unit with 3 sig figs.) (c) the masses (in grams) of CaF2 and H2SO4 that remain unreacted a the completion of the Determine (State your answers in kg unit with 2 decimal places.) reaction described in part (a)? Mass of CaF2 remai Mass of H SO rehainin Cypress College Fall 2016 CHEM 111AC (Gotoh) Exam 1 Page 5 of 10

Explanation / Answer

Balanced equation:
CaF2 + H2SO4 ===> CaSO4 + 2 HF

6 kg or 6000 gm of CaF2 = 6000 / 78.07 = 76.84 Moles

7.3 kg or 7300 gm of H2SO4 = 7300 /98.07 = 74.430 Moles

Moles of HF to be produced = 74.43 x 2 = 148.86 Moles

Limiting reagent is H2SO4

Excess reagent is CaF2

Theoretical yield of HF = 148.86 x 20 = 2978.15 gm or 2.978 Kg

Percentage yield = 2.86 x 100 / 2.978 = 96.04%

Excess reagent =6.0 - 5.811 = 0.189 Kg or 189 gm of CaF2

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