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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...

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

How many grams of reactant remain (excess) after 2.5 g sucrose and 25.0 g KClO3 react?

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Answer #1

C12H220, ($)+ 8 KClO3(s) → (260,69) + 114,069) +8 K Celsj Here, mom of sucrose : 2.5g 4 moles of sucrosemar mass-media Imol =

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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...
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