Question

Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation

C6H12O6(aq) + 6O2(g) — 6CO2(g) + 6H2O(1) 

Calculate the number of grams of oxygen required to convert 43.0 g of glucose to CO2, and H2O. 

mass of O2 = _______ g 

Calculate the number of grams of CO2, produced. 

mass of CO2 = _______ g 

Answer 1

1)

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 43 g

mol of C6H12O6 = (mass)/(molar mass)

= 43/1.802*10^2

= 0.2387 mol

According to balanced equation

mol of O2 required = (6/1)* moles of C6H12O6

= (6/1)*0.2387

= 1.432 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 1.432*32

= 45.83 g

Answer: 45.8 g

2)

According to balanced equation

mol of CO2 formed = (6/1)* moles of C6H12O6

= (6/1)*0.2387

= 1.432 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 1.432*44.01

= 63.03 g

Answer: 63.0 g

Answer 2

C6H12O6 (aq) + 69 (9) 6co(g) + 6H₂O (1) Molecular weight of Hyz 05 = 180 gm./mol Molecular weight of Oz -32 gmi mol Molecular weight of co2 = 44 gm/mol So, 1 mole of Co Hi 2 of reacts with 6 moles of a molecule and produce 6 moles of co, & 6 molesof to So, 180 gm of C6H6206 reacts with (32 x 60gm Oz and produce (44*6) gms of coz and (6x18)gms.of ho - igm of CG Hopo will react with = 32X6 gm of Oz . So, 43.0 gm of & H12Of will react with a 32x6x43 gm of a 180

= 45. 84 8m @ Mass of o, = 45 - 81 эт. 180 gm. c6 Hz of reacts with (3246)gmę nd produce (44x6) gm co , т а т с 2 б ю:u pro duce = 44 9m Co, A3 m. CH12O6 Jiu pro di co - 44xс x 43 = 63. 07 Яm. lgo • Mass of co= 63.07 gm.