Propane undergoes the following complete combustion reaction...

Propane undergoes the following complete combustion reaction: C3H8 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (g) + 2219 kJ What mass of propane must undergo combustion to convert 1,000.0 g of water at 20.0 °C to steam (gas) at 100.0 °C? The molar enthalpy of vaporization for water is 40.65 kJ/mol.

Transcript

00:01 Let's discuss this question.

00:02 So here propane undergoes the following combustion reaction.

00:06 So we need to find the mass of propane that must undergo combustion to convert 1 ,000 grams of water at 20 degrees celsius to steam at 100 degrees celsius.

00:16 So here we can say mass of water is equal to 1 ,000 grams and molar mass is equal to 18 grams.

00:33 So here we can say molar enthalpy of vaporization for water, that is delta h0v is equal to 40 .65 kilojune per mole.

01:00 So if we convert it into jule per mole we will be having this is equal to 40 .65 divided by 18 multiplied by 1 ,000 joule per grams.

01:14 So here, therefore delta h0v is equal to 2258 .0 .3 june per grams and this is the final answer.

01:27 Now further here we can say the amount of heat required to convert water from 20 degrees celsius to 100 degrees celsius.

01:53 This cp multiplied by m multiplied by delta t.

01:56 This is equal to 4 .183 multiplied by 1000 multiplied by 100 minus 20 so this is equal to 3344 4 .640 jule so here we can say this will be equal to 334 .64 .64 kilojoules there as here we can say cp is equal to 4 .183 jule per gram degree celsius so here we can say therefore amount of heat energy required to convert water at 100 degrees celsius into steam will be equal to m delta h0v so here we can say this is equal to thousand multiplied by 2258 .3 jule so this is equal to 2258 .3 kilojoules.

03:18 Amount of heat required to convert thousand grams of water at 20 degrees celsius to 100 degrees celsius strain will be equal to cp multiplied by m multiplied by delta t plus m multiplied by delta h0v so here we can say this will be equal to 2592 .94 kilojoules so here we can say further that the equation for combustion of propane will be c3 h8 plus 5 o2 that gives us 3 c o2 plus 4 h2o plus 2219 kilojoules this is the h value.

04:40 So here we can say according to this equation 1 mole is equal to 44 grams of c3h8 reduced.

04:53 So 2 to 19 kilojoule of heat on cumberson.

05:03 So here we can say there for the mass of propane required...

Question

Propane undergoes the following complete combustion reaction: C3H8 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (g) + 2219 kJ What mass of propane must undergo combustion to convert 1,000.0 g of water at 20.0 °C to steam (gas) at 100.0 °C? The molar enthalpy of vaporization for water is 40.65 kJ/mol.

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