# At constant pressure, the combustion of 5.00 g of C2H6 (g) releases 259 kJ of heat. What is ΔH for the reaction? 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

Mar 29, 2015

The heat of combustion is -3120 kJ.

You know that 5.00 g of C₂H₆ releases 259 kJ of heat.

You want to calculate the heat liberated by 2 mol of C₂H₆.

2 cancel("mol C₂H₆") × "30.07 g C₂H₆"/(1 cancel("mol C₂H₆")) = "60.14 g C₂H₆"

ΔH = 60.14 cancel("g C₂H₆") × "-259 kJ"/(5.00 cancel("g C₂H₆")) = "-3120 kJ"

Mar 29, 2015

The enthalpy change of combustion for that reaction will be $\text{-3120 kJ}$.

$2 {C}_{2} {H}_{6 \left(g\right)} + 7 {O}_{2 \left(g\right)} \to 4 C {O}_{2 \left(g\right)} + 6 {H}_{2} {O}_{\left(l\right)}$

Notice that 2 moles of ethane take part in this reaction.

Now, you were given the heat released when 5.00 g of ethane undergoes combustion. Use ethane's molar mass to determine how many moles of ethane must undergo combustion for that much heat to be released

$\text{5.00"cancel("g") * "1 mole"/("30.07"cancel("g")) = "0.1663 moles}$

Because the given reaction requires the combustion of 2 moles of ethane, the amount of heat released will be

2cancel("moles") * ("-259 kJ")/(0.1663cancel("moles")) = "-3114.85 kJ"

Rounded to three sig figs, the number of sig figs given for 5.00 g and 259 kJ, answer will be

$\Delta {H}_{\text{comb") = color(green)("-3120 kJ}}$