Please help!! Aluminum Boiling point!?

Question

Aluminum metal melts at 660 Celsius and boils at 2450 Celsius. Its density is 2.70 grams/cc.

Calculate the following:
A.) Assuming an ideal gas, the volume occupied by 20 kilograms of aluminum vapor at its boiling points and under pressure of 2280 torr.
B.) The volume occupied by 1000 kilograms of the metal at 25 Celsius.
C.) If the specific heat capacity of aluminum metal is 0.89 Joules per gram-degree C, how many calories of energy are required to heat 10 kilomoles of aluminum from 250 Celsius to 500 Celsius.
D.) If 13.5 grams of aluminum were dissolved in an excess of concentrated HCL, what volume of hydrogen gas would be released at STP? You may assume that the ideal gas law holds and that the vapor pressure of water of water is negligible compared to the pressure of the released hydrogen.

Please help! I need to see it step by step! I know i need to use the ideal gas law, but im still confused on where to start?

Answer 1

A) The atomic mass of Al is 26.98 gm/mol, so 20 kg of Al consists of

(20*10^3 gm)/(26.98 gm/mol) = 741.29 mol

The ideal gas law is:

P*v = n*R*T

where P is the pressure
V is the volume
R is the universal gas constant = 62.3637 (liter*torr)/(mol*K)
T is the absolute temperature
n is the number of moles

So, V = n*R*T/P

V = (741.29 mol)*(62.3637 liter*torr/(mol*K))*(2450+273.15 K)/(2280 torr)

V = 55214.9 liters

---------

B) Assuming the density you were given (2.7 gm/cm^3) is for a temperature of 25 C, then the volume occupied by 1000 kg of Al at 25 C would be:

(1000*10^3 gm)/(2.7 gm/cm^3) = 3.704*10^5 cm^3 = 0.3704 m^3


---------

C) 10 kilomoles of Al equals

(10*10^3 mol)*26.98 gm/mol) = 269.8 kg = 2.698*10^5 gm Al

Assuming the heat capacity is a constant over the temperature range 250 - 500 C, then the heat required to raise the temperature of 10 kilomoles of Al from 250 to 500 C is:

Q = (2.698*10^5 gm) (500C - 250C) * 0.89 J/(gm*C) = 6.003*10^7 J = 60.03 MJ

----------

D) The net reaction for the dissolution of Al metal in a solution of HCl is:

Al(s) + 3H+(aq) = Al3+(aq) + 1.5H2(g)

13 gm of Al is

(13 gm)/(26.98 gm/mol) = 0.482 mol Al

each mole of Al produces 1.5 moles of H2 gas, so the dissolution of 0.482 moles of Al would produce 1.5 * 0.482 mol = 0.723 mol H2.

You should know that at STP, one mole of an ideal gas occupies 22.4 liters, so 0.723 mol of H2 would occupy:

22.4 liter * 0.723 = 16.20 liters

If you didn't know that 1 mole og ideal gas occupies 22.4 liters at STP, then you can use the ideal gas law to calculate the volume:

V = (0.723 mol)*(62.3637 liter*torr/mol*K)*(273.15 K)/(760 torr)

V = 16.195 liters

Note that if you plug in 1 mol instead of 0.723 mol in the above equation and calculate the volume, you will prove that 1 mole of an ideal gas occupies a volume of 22.414 liters.

Answer 2

Don't panic!

A: get the parameters in proper units:

2280torr/760torr/atm = 3atm
2450 C + 273 = 2723K
[20kg x 1000g/kg]/27g/mol = 740.7moles

Plug into gas law (solved for volume; use proper R value)

v = [740.7 x 0.08205 x 2723]/3 = 55160 liters

B: density for aluminum = 2.7g/cc or 2.7kg/liter

1000kg/2.7kg/liter = 370.4 liters

Volume unit was not specified, but 1 liter = 1 dm3 so could easliy be converted to cubic meters etc.

C: 10 kmoles = 10000moles x 27g/mol = 270000g

[500-250 C] x 270000g x 0.89 joule/g C x 0.2389cal/joule

= 1.435 x 10^7cal

Balance the equation for Al dissolving in HCl, (this gives the ratio of reactants used to products formed):

2 Al(s) + 6 HCl(aq) > 2 AlCl3(aq) + 3 H2 (g)

13.5g/27g/mol = 0.5mol Al (would evolve 0.5/2 x 3 mole H2)

1 mole of any gas is 22.4 liters @STP; so 3/4 x 22.4 liters = 16.8 liters

I hope these are all ok.