# What mass of butane in grams is necessary to produce 1.5 × 10^3 kJ of heat? What mass of CO2 is produced? C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(g) ΔH rxn = –2658 kJ

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### 1 Answer

The change in enthaply of the reaction (delta H) is -2658 kJ. This means that 2658 kJ of heat is released from the exothermic reaction. From the chemical equation, we see that this amount is for a single mole of butane since the coefficient in front of the butane (C4H10) is 1. So for each mole of butane combusted, 2658 kJ of heat is released. So now let's start with 1.5 x 10^3 kJ, which is the same as 1500 kJ:

1500 kJ * (1 mole/2658 kJ) = 0.564 moles butane

Now convert this to grams:

0.564 moles * (58 g/1 mole) = 32.7 grams of butane

To find the mass of CO2 produced, we see from the chemical equation that 1 mole of butane produces 4 moles of CO2. So multiply the number of moles by 4 and then convert to grams:

0.564 moles * (4 mole CO2/1 mole butane) * (44 g/mol) = 99.3 grams CO2

**So the answers are 32.7 grams of butane are needed and 99.3 grams of CO2 are produced.**