Potassium chlorate (KCIO_3) oxidizes sucrose (C_12H_22O_11) according to the fol

Question

Potassium chlorate (KCIO_3) oxidizes sucrose (C_12H_22O_11) according to the following unbalanced reaction. What are the coefficients in the balanced chemical equation in the order in which it is written? KCIO_3(s)+ C_12H_22O_11(s)rightarrow KC1(s) + CO_2(g) H_20(e) 4.1.4, 12.11 8, 1, 8, 12, 11 4.1.4.12.22 8, 2, 8, 12, 11 2, 1, 2, 12, 11 Hydrazine (N_2K_4) can decompose to form nitrogen gas and ammonia gas. What is the sum of the coefficients in the balanced chemical equation? 4 7 5 8 6 Glucose (C_6H_12O_6) is oxidized by molecular oxygen to carbon dioxide and water. How many O_2 molecules are needed for each molecule of glucose that is oxidized? The average adult exhales about 1.0 kg of carbon dioxide each day. How much oxygen is needed in metabolizing glucose (C_6H_12O_6, 180 g/mol) to make that much carbon dioxide? 180 g 730 g 1800 g 1500 g 360 g The average adult exhales about 1.0 kg of carbon dioxide each day. How much glucose (C_6H_12O_6. 180 g/mol) is consumed to make that much carbon dioxide? 3.8 kg 0.68 kg 4.1kg 0.18 kg 0.43 kg

Explanation / Answer

39) 8 KClO3 + C12H22O11 -------> 8 KCl + 12 CO2 + 11 H2O
answer : d
40) 3N2H4 -----> N2 + 4NH3 = 3+1+4 = 8
answer : e
41) C6H12O6 + 6 O2 ------> 6 CO2 + 6 H2O
Answer : c
42) number of moles of CO2 = 1000/44 = 22.727 moles
number of moles of O2 required = 22.727 moles
mass of O2 = 22.727*32 = 727.264 g nearly equal to 730 g
answer : d
43) number of moles of CO2 = 1000/44 = 22.727 moles
number of moles of glucose required = 22.727/6 moles

                                   = 3.7878 moles
mass of glucose = 3.7878*180 = 681.8 g = 0.68 kg
answer : d

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