Using the following balanced chemical equation, determine the following 2AI(NO_3

Question

Using the following balanced chemical equation, determine the following 2AI(NO_3)_3 (aq) + 3Na_2 CO_3 (aq) rightarrow 6NaNO_3 (aq) + Al_2 (CO_3)_3 (s) a. The mass of sodium carbonate needed to make 34.5 g of aluminum carbonate. b. The mass of sodium nitrate produced from 2.356 g of aluminum nitrate. c. The mass of aluminum nitrate needed to produce 4.50 g of aluminum carbonate. d. The mass of aluminum carbonate produced from 4.50 g of aluminum nitrate. e. The total number of atoms reacted when 3.45 g of aluminum carbonate is produced. f. The total number of atoms reacted when 3.45 g of sodium nitrate is produced. g. The total number of atoms produced when 3.45 g of sodium carbonate is reacted. h. The total number of atoms produced when 3.45 g of aluminum nitrate is reacted.

Explanation / Answer

Balanced equation:
2 Al(NO3)3 + 3 Na2CO3 ====> 6 NaNO3 + Al2(CO3)3

Molar mass of Al(NO3)3 = 212.9962386

Molar mass of Na2CO3 = 105.98843856

Molar mass of NaNO3 = 84.99466928

Molar mass of Al2(CO3) = 233.9897772

Question a )

34.5 gm aluminium carbonate = 34.5 / 233.989 = 0.1474 Moles

Mass of Na2CO3 needed = 0.1474 x 3 x 105.988 = 46.88 gm

Question b

2.356 gm of Al(NO3) = 2.356 /212.99 = 0.0110 Moles

Mass of NaNO3 to be produced = 0.0110 x 3 x 84.99 = 2.82 gm

Question c

4.5 gm of Al2(CO3)3 = 4.5 / 233.989 = 0.019231 Moles

Mass of aluminium nitrate needed = 0.019231 x 2 x 212.99 = 8.192 gm

Question d

4.5 gm of Aluminium nitrate = 4.5 / 212.99 = 0.02112 Moles

Mass of Al2(CO3)3 to be produced = 0.02112 x 0.5 x 233.989 = 2.47 gm

Question e

3.45 gm Al2(CO3)3 = 3.45 / 233.989 =   0.01474 Moles

Moles of aluminium nitrate = 0.01474 x 2 = 0.02948 Moles

Moles of sodium carbonate = 0.01474 x 3 = 0.0442 Moles

Total no of atoms = (13 x 0.02948 + 6 x 0.0442 ) 6.023 x 1023 = 3.9 x 1023 atoms

Question f )

3.45 gm of sodium nitrate = 3.45 / 84.99 = 0.04059 Moles

Total no of atoms =   (13 x 0.01353 + 6 x 0.02029 ) 6.023 x 1023 = 1.79 x 1023 atoms

Question g

3.45 gm of sodium carbonate = 3.45 / 105.98 = 0.0325 Moles

Total no of atoms =   (5 x 0.06510 + 14 x 0.01085 ) 6.023 x 1023 = 2.87 x 1023 atoms

Question h

3.45 gm Aluminum nitrate = 3.45/212.99 = 0.016197 Moles

Total no of atoms =   (5 x 0.04859 + 14 x 0.00809 ) 6.023 x 1023 = 2.14 x 1023 atoms

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