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Connect https://newconnect.mheducation.com/flow/connect.html y q: Outlook Blackboard N Netflix e If 100 G Of Steam Co WebAss Saved 2 attempts loft Check my work Be sure to answer all parts. One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 669 mL of toluene (d= 0.867 g/ml.) is consumed when a sample of gasoline burns in air. (a) How many grams of oxya 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? × 10GA OD molecules (Enter your answer in scientific notation.) Prev 1 of 24 Next

Explanation / Answer

voliume of toluene= 6.69ml, density of toluene= 0.867 g/ml

mass of toluene= volume* density = 6.69 ml* 0.867 g/ml =5.8 gm

atomic weights : C=12 and H= 1

molar mass of toluene (C6H5CH3)= 7*12+8*1= 92 g/mole, moles of toluene =mass/molar mass= 5.8/92 =0.063 moles

the combustion of toluene is C6H5CH3+ 9O2 ---------->7CO2+ 4H2O

As per the stoichiometry of the reaction, 1 mole of toluene requires 9 moles of oxygen for complete combustion.

hence moles of oxygen required= 9*0.063=0.567 moles, mass of oxygen required= moles* molar mass =0.567*32 =18.144 gm

moles of gaseous products are based on moles of toluene. 1 mole of toluene gives total of 11 moles of gaseous products containing 7 moles of CO2 and 4 moles of water.

0.063 moles of toluene gives 0.063*11=0.693 moles of gaseous products.

1 mole of toluene gives 4 moles of water vapor

0.063 moles of toluene gives 4*0.063= 0.252 moles of water vapor

1mole of water vapor contains 6.023*1023 molecules

0.252 moles of water vapor contains 6.023*0.252*1023 molecules =1.52*1023 molecules

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