Complete combustion of a 0.70-mol sample of a hydrocarbon, CxHy, gives 4.20 mol of CO2 and 3.50 mol of H20. The molecular formula of the original hydrocarbon is

Md Ariful Islam

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0.7 CxHy + xO2 =4.2CO2 + 3.5H2O /0.7 CxHy + x/0.7O2 =6CO2 + 5H2O C6H10
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Combustion of hydrocarbon in a limited supply of air produces CO in addition to CO2 and h2o when 1.31 g of particular hydrocarbon was burned in air ,1.26 g CO ,1.98 g CO2

Molar mass of CO = 28.011.26 = 0.045mMolar mass of CO2= 44.011.98/44.01 = 0.045mMolar mass of hydrocarbon = 28.06 mol1.31/28.06 = 0.047m

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Complete analysis of flue gas if the hydrocarbon is burned with theoretical air and combustion is complete

Air contains 21 mol percent O2 and 79 mol percent of N2. The minimum amount of air which supplies the required amount of oxygen for complete combustion of a fuel...

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The standard enthalpy of the reaction: CO(NH2)24) + H20) -› CO2(g) + 2 NHa(a) is -133.42 kJ/mol. Calculate the standard enthalpy of formation of urea, CO(NH2)2(6).

Standard enthalpy of formation=-362kJ/mol

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128 grams of oxygen reacted. After the combustion of an unknown hydrocarbon, the total pressure of a 18.91-L container is 7.3347 atm at 22.7 degrees celsius. What is the hydrocarbon?

PV=nRT n1=PV/RT n=7.3347*101.3*18.91/8.31*295.7=5.718 moles n2=128/32=4 CO2 is gas at this condition CxHy + (x+y/4)O2=xCO2 +y/2H2O as we see,coefficient near O2 (x+y/4) must be biger that near CO2 x.Coefficients are proportional...

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Define enthalpy of combustion. Calculate the enthalpy of combustion of methanol if the enthalpies of formation of CO2,H20 andCH3OH are respectively . -393.5kj/mol,-285.8kj/mol and -239.0kj/mol.

Answer: Enthalpy of combustion is the heat released during combustion of one mol amount of any substance by oxygen.You can calculate the combustion enthalpy of methanol using the enthalpy...

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Balance the following reactions of combustion a) C2H4 (g) + O2 yields CO2(g) +H20(g) + Energy The energy is throwing me off from doing a normal balancing equation. Can you help me with this

C2H4(g) + 3 O2 → 2 CO2(g) + 2 H20(g) + Energy “Energy” means that the reaction is exothermic and the heat (energy) is released. There is no need to...

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Complete combustion of 8.10 g of a hydrocarbon produced 26.0 g of CO2 and 8.88 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

CxHy + O2= xCO2 + y/2H2O8.1. 26. 8.8812 x 26/44 = 7.2 g (C)2 x 8.88/18= 0.9 g (H)7.2+0.9= 8.17.2/12 : 0.9:1= 0.6:0.9= 1:1.5 = 4:6C4H6.

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A hydrocarbon Y on combustion gives 0.352g of carbon (IV)oxide and 0.18g of water .the mass of one mole of Y is 58g. (i)Determine the molecular formula of Y

C4H10

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An unknown compound containing only C and H was burnt, yielding 8.8g of CO2 and 4.5g of H20. With a molecular weight of about 58, what is the molecular formula?

Butane C4H10

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A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be A. C6H6 B. C4H10 C. C5H10 D. C4H6 Please show working

CxHy + O2 = xCO2 + y/2 H2OIf we burn 1 mol of certain hydrocarbon, we would get x mol of CO2 and y/2 mol of H2O.During combustion, we received...

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