Question

Propane, C3H8, liquefies under modest pressure, allowing a large amount to be stored in a container. Calculate the number of moles of propane gas in a 100 −L container at 3.00 atm and 27 ∘C. Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is 0.590 g/mL.

Answer 1

Answer:

The amount of moles of propane gas is 12.16 moles.

The amount of moles of liquid propane is 1337.87 moles.

Explanation:

We know that the container has 100 L at 3 atm and 27°C.

We can solve the problem by using the Gas Law formula:

$P \ V = n \ R \ T$

The number of moles is given:

$n = \frac{PV}{RT}\\n = \frac{3atm \times 100 L}{0.0822 \frac{L \ atm}{mol \ K}  \times (273.15+27K)}\\n= 12.16 moles$

If we have the same container of liquid propane we can calculate the mass of propane with density:

$\delta = \frac{M}{V}$

The mass is given by:

$M = \delta \times V \\M= 0.590 \frac{g}{ml}  \times \frac{1000 ml}{1 L} \times 100L= 59000 g$

The molar mass of propane is 44.1 g/mol so the number of moles of liquid propane is:

$M= \frac{m}{MM} =\frac{59000g}{44.1 g/mol} =1337.87 moles$

Answer 2

The number of moles is "0.1218 and 1337.87".

Moles calculation:

Using formula:

$\to n = \frac{PV}{RT}$

calculation:

For the first question:

$n= 3.00 \ atm \times \frac{K\times mol}{0.08206\ L \times atm} \times \frac{1.00 L}{300\ K}$

  $= 1.00 \ atm \times \frac{K\times mol}{0.08206\ L \times atm} \times \frac{1.00 L}{100\ K} \\\\= 1.00 \ atm \times \frac{K\times mol}{8.206\ L \times atm} \times \frac{1.00 L}{ 1 K} \\\\= 0.1218 \ moles$

For the second question:

$\to \frac{0.590\ g}{1 \ ml} \times 1.00 \times 10^3\ ml \times \frac{1\ mol CH_3H_8}{ 44.094\ g}$

by solving the value we get

$\to 1337.87 \ moles$

Therefore, the number of moles are "0.1218 and 1337.87".

Find out more information about the moles here:

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