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?
Homework Answers
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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...
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,...
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?
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)+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?
KClO3 + C12H22O11 = KCl + CO2 + H2O If 850.0 g of Potassium Chlorate is...
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C12H22O11 + 8 KClO3 12 CO2 + 11H2O + 8 KCl How many grams of reactant...
C12H22O11 + 8 KClO3 12 CO2 + 11H2O + 8 KCl How many grams of reactant remain in excess after 3 grams of C12H22O11 and 18 grams of KClO3 react? Show all steps and work.
Consider the reaction C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O...
Consider the reaction C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l) in which 10.3 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/oC (including its water). The temperature inside the calorimeter was found to increase by 21.7 oC. Based on this information, what is the heat of this reaction per mole of sucrose? Enter your answer numerically, in terms of kJ/mol and to three significant figures.
A mixture containing KClO3, K2CO3, KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according...
A mixture containing KClO3, K2CO3, KHCO3, and KCl was heated, producing CO2,O2, and H2O gases according to the following equations: 2KClO3(s)2KHCO3(s)K2CO3(s)→→→2KCl(s)+3O2(g)K2O(s)+H2O(g)+2CO2(g)K2O(s)+CO2(g) The KCl does not react under the conditions of the reaction. 100.0 g of the mixture produces 1.70 g of H2O, 12.66 g of CO2, and 4.00 g of O2. (Assume complete decomposition of the mixture.) You may want to reference (Page) Section 3.7 while completing this problem. a. How many grams of KClO3 were in the original mixture?...
The nonvolatile, nonelectrolyte sucrose , C12H22O11 (342.3 g/mol), is soluble in water H2O. How many grams...
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