C12H22O11(s) + 8 KClO3(s) -> 12 CO2(g) + 11 H2O(g) + 8 KCl(s) KClO3 = 122.6...

Question

Question

C12H22O11(s) + 8 KClO3(s) -> 12 CO2(g) + 11 H2O(g) + 8 KCl(s)

KClO3 = 122.6 g/mol

sucrose = 342.3 g/mol

CO2 = 44.01 g/mol

1 skittle = 0.84 grams of sucrose

** 17.85 (g) remaining of KCLO3- reactant

** 5 skittles from sucrose

If 2.2 grams of CO2 are produced, what is the percent yield?

Answer 1

Answer #1

Solution CH₂ 1 + 8 Kuo 12 co + 11 Hot 8 KG Moles of sucrose 2) 5x084g 122.342.3 g. moll 0.012 moles 342.3 # # Sucrose is limi

Answer 2

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