Question

Please help me with this question

Answer 1

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. $Na_2SO_3 + Ca(NO_3)_2 --- > 2NaNO_3 + CaSO_3$

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.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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