On this page are two \"show your work\" questions. On the exams \"show your work

Question

On this page are two "show your work" questions. On the exams "show your work questions" can receive partial marks if you were able to do some part of the question correctly. 15. Chang 3.89. .00 kg calcium fluoride is reacted with cxcess sulfuric acid. The reaction yields 2.86 kg HF. Balanced reaction: CaF2 + H1SOs CaSO4 + 2HF Calculate the % yield of the reaction. 16. Chang 3.86 (but changed from Chang's version) 48.2 g HCl is added to 74.7 g Mn02. The balanced reaction is: MnOs + 4HC1 MnCh + Cl2 + 2 H2O a) How much (g) Clz can be produced? b) How much (g) of the excess reactant will be left over?

Explanation / Answer

molar mass of CaF2 = 78 gm/mol

6 kg CaF2 = 6000 /72 = 83.3 mole

mole of HF produce = 2*83.3 = 166.6 mole

weight of HF = 20*166.6 = 3332 gm = 3.332kg

% yield = (2.86/3.332)*100 = 85.8%

mole of HCL = 48.2/36.5 = 1.32 mole

mole of MnO2 = 74.7 / 87 =0.86 mole

limiting reagent = HCl

so, mole of Cl2 will be = (1/4)*1.32= 0.33 mole

weight of Cl2 = 71 * 0.33 = 23.43 gm

mole of remaining MnO2 = 0.86 - 0.33 = 0.53 mole

weight of 0.53 mole of MnO2 = 87*0.53= 46.11 gm (answer)

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