what product of the combustion of a hydrocarbon occurs only when the supply of oxygen is limited? a. sulfur dioxide b. nitrogen dioxide c. carbon dioxide d. carbon monoxide

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c. carbon dioxide

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What mass of sulfur trioxide is produced from the reaction of 11.4 grams of sulfur dioxide and 3.15 grams of oxygen?

Mass of sulfur trioxide is produced=7.2g

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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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If liquid carbon disulfide reacts with 450 mL of oxygen to produce to gases carbon dioxide and sulfur dioxide, what volume of each product is produced? a. 20 mL b. 150 mL c. 0.150 L d. 450 L

150 mL

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10cm3 of a gaseous hydrocarbon C2 Hx required 30cm3 of oxygen gas for complete combustion if steam and 20cm3 of carbon iv oxide produced what is the value of x

First we can write the chemical equation of hydrocarbon burning:C2Hx+(2+0.25x)O2=2CO2+0.5xH2OThen we can see that volumes of hydrocarbon and oxygen are related as 1:3. We can make an equation:2+0.25x=3x=4Answer: x=4

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In the reaction, 150. g of ammonia (NH<sub>3</sub>) were reacted with 150. g of oxygen (O<sub>2</sub>) to produce nitrogen monoxide (NO). How much nitrogen monoxide would be produced?

Mass of Nitrogen monoxide =73.5g

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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

0.7 CxHy + xO2 =4.2CO2 + 3.5H2O /0.7 CxHy + x/0.7O2 =6CO2 + 5H2O C6H10

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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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Four moles of carbon monoxide (CO) has 48 grams of carbon (C) and 64 grams of oxygen (O). What is the percent composition of carbon in carbon monoxide?

Answer: m (CO) = n*M, where n is the amount of substance, mole, M is the molar mass, g/mole m (CO) = 4*28 = 112 g % C = m...

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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 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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