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

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

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

1. If each skittle contains 0.84 g of sucrose (C12H22O11)

What is the minimum number of skittles needed to maximize the product (no KClO3 remaining) if I use 12.0 g of KClO3?

science chemistry

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Answer 1

Answer #1

Given that: 6127220, C3) + Skeloses) - 12 C02 103 +114,067 8icles Mole ratio: I more sucrose - smols kelos. More (n) = Weight - Noot skittle = weight of sucrose = Skittle 4.180g 0.848 E 4.983.7 the Numbes of skittles 25

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Answer 2

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

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