Question

I need answer for the A,B and C parts.

3 out of 3 attempts One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 0.325 mL of toluene (d 0.867 g/mL) is consumed when a sample of gasoline burns L of toluene (d -0.867 g/mL) is consumed when a sample of gasoline burns in air. (a) How many grams of oxygen are needed for complete combustion of the toluene? (b) How many total moles of gaseous products form? mol (e) How many molecules of water vapor form?

Answer 1

Solution: If toluene having density value of 0.867 g/mL, undergoes combustion reaction, then it reacts as, i . ii. 2 C6H3CH3 (1) + 18 02 (g) = 14 CO2 (g) + 8 H2O (8) If 0.325 mL of toluene is consumed, then the mass of toluene is used in the process = (0.325 mL < 0.867 g/mL) = 0.282 g a. From the balanced reaction of the process, we see that 2 moles of toluene react completely with 18 moles of oxygen. The molar mass of toluene is 92.14 g/mol which means, 92.14 gm of tolune is equivalent to 1 mole 1 gm of tolune is equivalent to 14 1 x0.282 gm 0.282 gm of tolune is equivalent to 92.14 gm (22.14) mole mole = 3x10-3 mole The ratio of toluene : oxygen = 2 : 18 = 1: 9 Therefore, 3x10-3 mole of toluene will completely react with (3x10 3 x 9) = 0.027 mole of oxygen Similarly, the molar mass of oxygen is 32 gm/mol So, 1 mole of O2 is equivalent to 32 gm 0.027 mole of O2 is equivalent to (32 x 0.027) gm = 0.864 gm Hence, for complete combustion of toluene, 0.864 gm of oxygen is required.

b. The ratio of toluene : carbon dioxide = 2:14 = 1: 7 Therefore, 3x10-3 mole of toluene will completely react to produce (3x10-3 x 7) = 0.021 mole of CO2 which is one of the gaseous products formed. Similarly, the ratio of toluene : water = 2: 8 = 1:4 Therefore, 3x10-3 mole of toluene will completely react to produce (3x103 x 4) = 0.012 moles of H2O which is the other gaseous product formed. So, total moles of gaseous products formed is (0.021 + 0.012) = 0.033 moles. C. From Avogadro's law, we know that 1 mole of H2O is equivalent to 6.023 x 1023 number of molecules Or, 0.012 mole of H2O is equivalent to (6.023 x 1023 x 0.012) = 7.227 x 1021 number of molecules So, number of water vapour molecules form is 7.227 x 1021