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

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

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= .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 if I use 12.0g of KClO3?

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

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

science chemistry

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Answer 1

Answer #1

Gitutat 2-0 2 22 MoSucose Moje Yofio Mo (n) MO Jeculuy hyt 12-9 122 6 Kclo Mol mol 0.09789 of kclos Molus Su eofos e Molus 0 No Si Sucdose 4.186 4.988 he Numb St s 5 Mo hsof 2.5 27 Suctose wiytotelog = 25.07 0 00 730 mos suctose Yuie 8 xo.00730 Mo e klo3 s0.05gy Moles MOles Moot kolog X Mok

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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 = 122.6...

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?
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?
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?

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

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

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