# How many grams of C2H4 must burn to give 84.0 kJ of heat?

Jul 11, 2017

Approx. $1.7 \cdot g$ of ethylene are required.........

#### Explanation:

We have the stoichiometric equation........

${H}_{2} C = C {H}_{2} \left(g\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta$

$\Delta {H}^{\circ} = - 1411 \cdot k J \cdot m o {l}^{-} 1$, and this means per mole of reaction as written.......And thus complete combustion of $28 \cdot g$ ethylene ($1 \cdot m o l$) results in the release of $1411 \cdot k J$ of energy (the minus sign means that energy is RELEASED; the reaction is exothermic). And we could use this energy to heat our homes, or generate electricity, or cook our breakfast.

We want $84 \cdot k J$ energy released, and so we solve the quotient....

(84*kJ)/(1411*kJ*mol^-1)xx28.01*g*mol^-1=??*g

And thus we have used $\Delta$ as a stoichiometric product of the reaction. We could likewise work out how much carbon dioxide and water were produced (and of course this bond FORMATION is repsonsible for the energy release). Capisce?