Nitromethane (CH 3 NO 2 ), sometimes used as a fuel for drag racing, burns according to the following reaction: 4 CH 3 NO 2 (l) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) + 4 NO 2 (g) ∆Hº = – 2441.6 kJ How much heat is released by burning 125.0 g of nitromethane (MW:61.044 g/mol)?

Here we're told that nitromethane, which is c h three n o two, sometimes used as a fuel for drag racing, burns according to the following reaction. Here, this reaction has an enthalpy of reaction equal to 2 thou negative 2441.6 kilojoules. Here they're saying, how much heat is released by burning a 125 grams of nitromethane. Here we're told that the molecular weight is 61.044 grams per mole. Alright. So here they're asking me to figure out the amount of heat released, so this is delta h, And they're giving me grams of nitromethane, so this is my grams of given. So what I need to do here is we need to use the grams of given initially, so we have a 125 grams of Nitromethane. We're gonna convert the grams of given into moles of given. So for every one mole of Nitromethane it weighs 61.044 grams. So So grams cancel out, now I have moles of nitromethane. Now because I have moles of nitromethane, I can determine the, delta h of unknown here. So we're gonna say on my according to my balanced equation, it's this much for my delta h. Every time there are 4 moles of Nitromethane. So that's the delta h to mole comparison. So now the moles cancel out and what I'll have left at the end are kilojoules. So here it's negative 249.918 kilojoules. Here, this has 5 sig figs, this has 4 sig figs, so we'll go with 4 sig figs. So that's negative 1.250 times 10 to the 3 kilojoules of heat are released when burning a 125 grams of nitromethane.

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7. Energy, Rate and Equilibrium

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GOB Chemistry

7. Energy, Rate and Equilibrium

Thermochemical Equations

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Multiple Choice

Nitromethane (CH₃NO₂), sometimes used as a fuel for drag racing, burns according to the following reaction:

4 CH₃NO₂ (l) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (g) + 4 NO₂ (g) ∆Hº = – 2441.6 kJ

How much heat is released by burning 125.0 g of nitromethane (MW:61.044 g/mol)?

1.250 x 10³ kJ

-1.250 x 10³ kJ

2.440 x 10³ kJ

-2.440 x 10³ kJ

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Verified step by step guidance

Determine the number of moles of nitromethane (CH3NO2) by using its mass and molar mass. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the mass is 125.0 g and the molar mass is 61.044 g/mol.

Calculate the moles of nitromethane: \( \text{moles of CH3NO2} = \frac{125.0 \text{ g}}{61.044 \text{ g/mol}} \).

Use the stoichiometry of the balanced chemical equation to find the relationship between moles of nitromethane and the enthalpy change (\( \Delta H^\circ \)). The reaction shows that 4 moles of CH3NO2 release 2441.6 kJ of heat.

Calculate the heat released for the moles of nitromethane you have. Use the proportion: \( \text{heat released} = \left( \frac{\text{moles of CH3NO2}}{4 \text{ moles}} \right) \times (-2441.6 \text{ kJ}) \).

Convert the calculated heat to the appropriate units and sign, ensuring it matches one of the given answer choices. Remember that the heat released should be a negative value, indicating an exothermic reaction.