1. The resulting concentration will be 0.00044 mol/L
2. The minimum mass of sodium sulfite to add will be 0.4032 grams.
Stoichiometric problems
1. Using m1v1=m2v2
m1 = 0.01 mol/L, v1 = 20 mL, v2 = 450 mL
m2 = m1v1/v2 = 0.01 x 20/450 = 0.00044 mol/L
2. [tex]Na_2SO_3 + Ca(NO_3)_2 --- > 2NaNO_3 + CaSO_3[/tex]
Mole ratio of the reactants = 1:1
Mole of 80 mL, 0.0400 mol/L Ca(NO3)2 = 80/1000 x 0.0400 = 0.0032 mol
Equivalent mole of Na2SO3 = 0.0032 moles
Mass of 0.0032 moles Na2SO3 = 0.0032 x 126 = 0.4032 grams
Thus, the minimum mass of sodium sulfite to be added must be 0.4032 grams.
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