When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass o
1 answer:
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>First, the equation of the reaction:
The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
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