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C12H22O11(s)+8KCLO3(s)=12CO2(g)+11H2O(g)+8KCL(8) 1. If each skittle contains 0.84 g of sucrose (C12H22O11) what is the m...

C12H22O11(s)+8KCLO3(s)=12CO2(g)+11H2O(g)+8KCL(8)

1. If each skittle contains 0.84 g of sucrose (C12H22O11) what is the maximum number of skittles needed to maximize the product (no KCLO3 remaining) if i use 12.0 g of KCLO3?

2. how many grams of reactants remains (excess) after 2.5 g sucrose and 25.0 g KCLO3 react?

3. if 2.2 gram of CO2 are produced what is the precent yield?

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Answer #1

Moles of Kelon = mass taken o Molar mas - 0./6/ 74.5 let n number of skittle is needed then total mass of sucose - 0.84ng so,

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C12H22O11(s)+8KCLO3(s)=12CO2(g)+11H2O(g)+8KCL(8) 1. If each skittle contains 0.84 g of sucrose (C12H22O11) what is the m...
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