Question

When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass of sodium sulphate that was used

Answer 1

The mass of sodium sulfite that was used will be 1,890 grams.

Stoichiometric problems

First, the equation of the reaction:

$NaSO_3 + 2HCl --- > NaCl_2 + H_2O + SO_2$

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

More on stoichiometric problems can be found here: brainly.com/question/14465605

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