3. Use the following chemical reaction to answer the questionsbelow: KClO3 + 6KB

Question

3. Use the following chemical reaction to answer the questionsbelow:
KClO3 + 6KBr + 3H2SO4 ? KCl + 3Br2 + 3H2O + 3K2SO4
a. How many moles of H2SO4 are needed to resct with 3.70 gKClO3?
b. If 305g of KClO3 and 415g of KBr are mixed with excess H2SO4,what is the theoretical yield?
c. If 234g Br2 is recovered, what is the percentage yield in thisreaction?

Explanation / Answer

first, you have to convert the initial g to moles. a) the KClO3 has Pm = 122.5 g/mol. 3.7 g --> 0.03 moles of yourcompound. if you know that for each mole of KClO3 you need 3 molesof H2SO4, for do the reaction you need 0.09moles b) convert all to moles, using the Pm. KBr Pm = 119 g/mol      KClO3 305 g --> 2.5 moles      KBr 415 g --> 3.5 moles. you cant use more KBr than what you have, this is the limitingreagent because for each mol of KClO3 you need 6 moles of KBr. youwill use 0.58 moles of KClO3 for making thereaction. and you will obtein 3 moles of Br2 for each mol of KClO3used, that is 1.74 moles of Br2 c) we must obtein 1.74 moles of Br2, convert this to g using Br Pm= 160 g/mol.     Br2 1.74 moles --> 278.4 g if we obtein 234 g at the end. R = 234/278.4 *100 -->R = 84%

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