Please help! Aluminum metal melts at 660 degrees C and boils at 2450 degrees C. Its density is 2.70 grams/cc.?

Calculate the following:


A. Assuming an ideal gas, the volume occupied by 20.0 kilograms of aluminum vapor at its boiling point and under a pressure of 2280 torr.


B. The volume occupied by 1000. kilograms of the metal at 25 degrees C.


C. If the specific heat capacity of aluminum metal 0.89 Joules per gram-degree C, how many calories of energy are required to heat 10 kilomoles of aluminum from 250 degrees C to 500 degrees C.


D. If 13.5 grams of aluminum were dissolved in an excess of concentrated HCI, 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 is negligible compared to the pressure of the released hydrogen.

Please help! Aluminum metal melts at 660 degrees C and boils at 2450 degrees C. Its density is 2.70 grams/cc.?
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.... K)/(2280 torr)





V = 55214.9 liters





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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








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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 (not a particularly good assumption -- the heat capacity is itself a function of temperature), 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





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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)





13.5 gm of Al is





(13.5 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 * 0.5mol = 0.75 mol H2.





You should know that at STP, one mole of an ideal gas occupies 22.4 liters, so 0.75 mol of H2 would occupy:





22.4 liter * 0.75 = 16.80 liters





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





V = (0.75 mol)*(62.3637 liter*torr/mol*K)*(273.15 K)/(760 torr)





V = 16.8 liters





Note that if you plug in 1 mol instead of 0.75 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.
Reply:A. 20,000g / 26.98g/mol = 741.3 mol


V= nRT/P= 741.3 mol * 8.314 J/molK * (2450+273)K / {2280torr / 760(torr/atm) * 101,325 (Pa/atm)} after all the stuff cancels, you get:


V=55.21 cubic meters





B. 1000 kg = 1,000,000 grams. Since it's solid at this T, use the solid density of 2.70 g/cc:


V=mass/density =1,000,000g/2.70g/cc = 370,400 cc or 370 liters





C. E=mc(delta T) = (10 Kmol )(0.89 J/gC)(500-250C) (1000 mol/Kmol)(26.98 g/mol) = 60,030,000 J = 60 MJ





D. 13.5 g / 26.98g/mol = 0.500 mol





The balanced equation is:


2Al + 6HCl ~%26gt; 2AlCl3 + 3H2


Assuming that all of aluminum reacts we should get (3/2)(0.500mol) = 0.75mol of H2 gas...(the 3/2 is from stoichiometry of Al to H2)





V = nRT/P = (0.75 mol) * (0.082057 L-atm/mol/K) * (273 K)/ 1atm = 16.8 L


or alternately, you can say that at STP, 1 mol of idea gas has volume of 22.4L, then 0.75 moles have volume of 16.8L