# Question c91cf

Oct 25, 2015

Here's what I got.

#### Explanation:

Your question is missing a very important piece of information - the enthalpy change of combustion for liquified petroleum gas, or LPG.

WIthout knowing how much heat is released when LPG undergoes combustion you cannot determine how much LPG would be needed to heat the water.

I'm assuming that the question given to you included a chemical equation. To keep things simple, you can represent LPG as either propane, ${\text{C"_3"H}}_{8}$, or butane, ${\text{C"_4"H}}_{10}$ (usually it's a mixture of the two).

I'll show you how to solve the problem for propane. The combustion reaction of propane is

${\text{C"_3"H"_text(8(g]) + 5"O"_text(2(g]) ->3"CO"_text(2(g]) + 4"H"_2"O}}_{\textrm{\left(l\right]}}$

The standard enthalpy change of combustion for propane is $\Delta {H}_{\text{comb"^@ = -"2202 kJ/mol}}$.

Here's you strategy for this problem. Use the given volume of water to determine how much heat would be needed to increase the water's temperature from ${25}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$.

The equation that establishes a relationship between heat and temperature increase looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed
$m$ - the mass of the water, equal to 4.184"J"/("g"""^@"C")
$c$ - the specific heat of water
$\Delta T$ - the change in temperature

SInce no information about water's density was given, you can assume it to be equal to $\text{1 kg/L}$.

This means that you have

1.5color(red)(cancel(color(black)("L"))) * (1color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("L")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.5 * 10^3"g"

The amount of heat needed to increase the temperature of the water will be

$q = 1.5 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.184"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (100 - 25)color(red)(cancel(color(black)(""^@"C}}}}$

$q = 470.7 \cdot {10}^{3} \text{J" = "470.7 kJ}$

Now look at the enthalpy of combustion for that reaction. If one mole of propane gives off $\text{2202 kJ}$, you would need

470.7color(red)(cancel(color(black)("kJ"))) * ("1 mole C"_3"H"_8)/(2202color(red)(cancel(color(black)("kJ")))) = "0.2138 moles C"_3"H"_8

to produce the amount of heat you need to heat the water.

However, keep in mind that this amount would be needed if 100% of the heat produced by the combustion reaction went to heat the water.

You know that that's not the case, since only 16% actually heats the gas.

This means that you would need more propane to make sure that you get this much heat

$0.2138 \cdot \frac{100}{16} = {\text{1.336 moles C"_3"H}}_{8}$

Finally, to get the mass of propane needed, use its molar mass

1.336color(red)(cancel(color(black)("moles"))) * "44.1 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("58.9 g C"_3"H"_8)#