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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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

First, determine the number of moles of nitromethane (CH3NO2) in 125.0 g. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of nitromethane is 61.044 g/mol.

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

Next, use the stoichiometry of the reaction to determine the heat released. The reaction shows that 4 moles of CH3NO2 release 2441.6 kJ of heat.

Set up a proportion to find the heat released by the moles of nitromethane calculated: \( \text{heat released} = \left( \frac{\text{moles of CH3NO2}}{4 \text{ moles}} \right) \times 2441.6 \text{ kJ} \).

Finally, ensure the sign of the heat released is negative, as the reaction is exothermic, meaning heat is released.