C12H22O11 + 8 KClO3 12 CO2 + 11H2O + 8 KCl How many grams of reactant...

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Question

C₁₂H₂₂O₁₁ + 8 KClO₃12 CO₂ + 11H₂O + 8 KCl

How many grams of reactant remain in excess after 3 grams of C₁₂H₂₂O₁₁ and 18 grams of KClO₃ react?

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Answer 1

Answer #1

C12 H₂2011 + 8k clo, → 1220₂ + 11120+8kce = 8.76X163 mod moles of C2 H2 2011 = 3g - 8.76mmal 342.299/mol moles of klo = 18g =

Answer 2

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