Question

What mass of precipitate is formed from the reaction of 12.00 mL of 0.600M aluminum nitrate with 11.50 mL of 0.850M sodium carbonate according to the unbalanced chemical reaction? What volume of excess reagent is left over? ____ Al(NO3)3 (aq) + ____ Na2CO3 (aq) → ____ NaNO3 (aq) + ____ Al2(CO3)3 (s)

Answer 1

2Al(NO3)3(aq)+3Na2CO3(aq) ------> 6NaNO3(aq) + Al2(CO3)3

Stoichiometric masses

Al(NO3)3 = 2mol × 212.996g/mol = 425.992g

Na2CO3 = 3mol × 105.988g/mol = 317.994g

NaNO3 = 6 mol × 84.995g/mol = 509.97g

Al2(CO3)3 = 1mol × 233.989g/mol = 233.989g

Actual masses

Al(NO3)3 = (0.6 mol /1000ml)×12ml)×212.996g/mol= 1.5336g

Na2CO3= (( 0.850mol/1000ml)×11.50ml )×105.998g/mol =1.0361g

According to stoichiometric masses calculation, 425.992g of Al(NO3)3 react with 317.994g of Na2CO3

Therefore, 1.5336g of Al(NO3)3 react with (317.994g/425.992g)× 1.5336g = 1.1448g of Na2CO3

Available mass of Na2CO3 = 1.0361g

Therefore, Na2CO3 is limiting

According to stoichiometric masses calculation , 233.989g of Al2(CO3)3 comes from 317.994g of Na2CO3

Therefore, 1.0361g of Na2CO3 give (233.989g/317.994g)×1.0361g = 0.7624g of Al2(CO3)3

Therefore,

Theoretical yield of Al2(CO3)3 = 0.7624g

Excess reagent is Al(NO3)3

1.0361g of Na2CO3 react with (425.992g/317.994g)×1.0361g = 1.3879g of Al(NO3)3

Therefore , excess mass of Al(NO3)3 left = 1.5336g - 1.3879g = 0.1457g