Consider the following reaction: 2 C 6 H 6 (l) + 15 O 2 (g) → 12 CO 2 (g) + 6 H 2 O (g) ∆Hº = – 6278 kJ What volume of benzene (C 6 H 6 , d = 0.880 g/mL, molar mass = 78.11 g/mol) is necessary to evolve 5.19 x 10 9 kJ of heat?

Here it says consider the following reaction. Here we have 2 moles of benzene reacting with 15 moles of oxygen gas to produce 12 moles of carbon dioxide gas, plus 6 moles of water vapor. Associated with it is our delta h of reaction as negative 6,278 kilojoules. Here it says, what volume of benzene, which is c 6h6, that has a density of 0.880 grams per milliliter and a molar mass of 78.11 grams per mole, is necessary to evolve 5.19 times 10 to the 9 kilojoules of heat. So first of all, this talking about heat, heat is q, but q is also equal to delta h. So here they're giving us delta h and they want me to determine the volume, so milliliters of benzene. So that's what our missing information is. We're gonna start out with 5.19 times 10 to the negative 9 kilojoules. Now realize from the information given to us that here is another kilojoule. We're gonna bring that down and we're gonna realize that that many kilojoules released is attached to 2 moles of benzene. So that's for every 2 moles of benzene. Kilojoules cancel out, now I have moles of benzene. I need to get to volume in milliliters of benzene, which means density is gonna have to be incorporated in some way. But density has within it grams. So what I need to do next is I need to convert moles of benzene into grams of benzene. One mole of benzene, we're told weighs 78.11 grams. Now that I have the grams of benzene, I can say that the density of benzene is 0.880 grams of benzene for every 1 milliliter of benzene. So grams cancel out and I'll have milliliters. Here I go I'm going to isolate 3 sig figs based on the values given to me. So this comes out to 7.34 times 10 to the 7 milliliters of benzene as the final volume for

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1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
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5. Classification & Balancing of Chemical Reactions3h 17m
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13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
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16. Carboxylic Acids and Their Derivatives1h 11m
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7. Energy, Rate and Equilibrium

Thermochemical Equations

GOB Chemistry

7. Energy, Rate and Equilibrium

Thermochemical Equations

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Multiple Choice

Consider the following reaction:
2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g) ∆Hº = – 6278 kJ
What volume of benzene (C₆H₆, d = 0.880 g/mL, molar mass = 78.11 g/mol) is necessary to evolve 5.19 x 10⁹ kJ of heat?

1.47 x 10⁸ mL

5.19 x 10⁹ mL

4.37 x 10⁷ mL

9.51 x 10⁹ mL

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Verified step by step guidance

Start by understanding the given reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) with ∆Hº = – 6278 kJ. This indicates that 6278 kJ of heat is released when 2 moles of benzene react.

Calculate the number of moles of benzene required to evolve 5.19 x 10^9 kJ of heat. Use the relationship: moles of C6H6 = (desired heat / heat per 2 moles of C6H6).

Convert the moles of benzene to grams using its molar mass (78.11 g/mol). Use the formula: mass (g) = moles x molar mass.

Convert the mass of benzene to volume using its density (0.880 g/mL). Use the formula: volume (mL) = mass (g) / density (g/mL).

Review the calculated volume and compare it with the given options to identify the correct answer.

1:13m

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Struggling with GOB Chemistry?

Consider the following reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2­O (g) ∆Hº = – 6278 kJWhat volume of benzene (C6H6, d = 0.880 g/mL, molar mass = 78.11 g/mol) is necessary to evolve 5.19 x 109 kJ of heat?

Watch next

Thermochemical Equations practice set

Consider the following reaction:
2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g) ∆Hº = – 6278 kJ
What volume of benzene (C₆H₆, d = 0.880 g/mL, molar mass = 78.11 g/mol) is necessary to evolve 5.19 x 10⁹ kJ of heat?