Question

33. Determine how many grams of carbon dioxide are produced in the oxidation of 0.00700 Kg of glucose, CH,206 CH,206(s) +602(8) → 6 CO2(g) + 6 H20(1) A.4.20 B. 7.20 C. 10.3 D. 2.76

Answer 1

CH120665) + 602.692 +6C0259) + 6H200 mass of glucose 0.00700 Kg molar mass of glucose = 6X12+12X1+6X 16 18001 mol = 1 kg 1000g ... mass of glucose = 79 moles of glucose = 79 [ me moles= mass 180 g/mol molar mass moles of glucose = 0.0388 Now i mole of glucose gives →6 mole of CO2 -0.0388 mole og glucose gives — 6X0.0338 mole of co COZ 1. moles of = 0.2328 mole

molax mass of CO2 = 1X12 +2416 12 +32 4 u glmol C C moles mass molar mass į. 0.2398 mol mass 44gimol mass of coa 0.2328.844 g mass of CO 2 10:3