Question

Given: Sodium Carbonate (Volume=2.00 L Mass of Chemical 3.947 g Temperature in Celcius 750 Pressure 1565.6 Torr) Aluminum Carbonate (Volume=1.50 L Mass of Chemical 2.123 g Temperature in Celcius 2200 Pressure 2134.3 Torr)

1. Write balanced equations for the composition of: Na2CO3(s) and Al2(CO3)3

2. Sample calculations for the number of moles og gas "n" produced in the decomposition of the compounds

3. Calculate Averaged Ideal Gas Constant.

Below are my calculations, I know they are incorrect because they are nowhere near the literary Idean Gas Constant which is 0.08206 atm-L/mol-K

2. Na2CO3 (s) --> Na2O(s) + CO2(g) MM= 105.99 (3.947 g Na2CO3 / 105.99g/mol Na2CO3) * (I mole Na2CO3 / 1 mole gas)= 0.037239 mol gas

Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g) MM= 233.99 (2.123 g Al2(CO3)3 / 233.99 g/mol Al2(CO3)3) *(1 mole Al2(CO3)3 / 3 mole gas)= 0.0030243 mol gas

3. R= PV/nT

Na2CO3 R = (1565.6 Torr * 1atm/760 Torr)(2.00 L) / (0.037239 moles)(750+273.15= 1023.15 K)

Al2(CO3)3 R = (2134.3 Torr * 1 atm/760 atm)(2.00 L) / (0.0030243)(2200 + 273.15= 2473.15 K)

I'm not sure what I am doing wrong. Thank you in advance!

Answer 1

Solution :-

1. Write balanced equations for the composition of: Na2CO3(s) and Al2(CO3)3

Solution :- balanced reaction equations

Na2CO3 (s) --> Na2O(s) + CO2(g)

Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g)

2. Sample calculations for the number of moles og gas "n" produced in the decomposition of the compounds

Solution :-

(3.947 g Na2CO3 / 105.99g/mol Na2CO3) * (1 mole gas/ 1 mole Na2CO3)= 0.037239 mol gas

(2.123 g Al2(CO3)3 / 233.99 g/mol Al2(CO3)3) *( 3 mole gas/1 mole Al2(CO3)3) = 0.027296 mol gas

3. Calculate Averaged Ideal Gas Constant.

Below are my calculations, I know they are incorrect because they are nowhere near the literary Idean Gas Constant which is 0.08206 atm-L/mol-K

PV= nRT

Calculate the R for the Na2CO3

R = (1565.6 Torr * 1atm/760 Torr)(2.00 L) / (0.037239 moles)(1023.15 K)

    = 0.108 L atm per mol K

Calculating R for the Al2(CO3)2

R= PV/nT

Al2(CO3)3 R = (2134.3 Torr * 1 atm/760 atm)(2.00 L) / (0.027296 mol)( 2473.15 K)

                          = 0.0624 Latm per mol K