How do you write an equation to represent Calcium sulfite decomposes to form calcium oxide and sulfur dioxide?

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    ##CaSO_text(3(s]) -> CaO_text((s]) + SO_text(2(g]) uarr##
    The first clue to go by is the fact that you’re dealing with a so you know that calcium sulfite is the only reactant and that you have two products calcium oxide and sulfur dioxide.
    Start by focusing on the reactant calcium sulfite.
    As you know the -ite suffix is used to distinguish between that contain oxygen. These polyatomic anions go by the name of oxyanions.
    More specifically the -ite suffix is used for that can form multiple oxyanions to represent an oxyanion that contains fewer atoms of oxygen.
    By comparison for the same element the -ate suffix is used to represent the oxyanion that contains more atoms of oxygen.
    In this case sulfite is the name given to the ##SO_3^(2-)## anion. Sulfate for example is the name given to the ##SO_4^(2-)## anion since it contains an extra oxygen atom than the sulfite anion.
    So you know that the reactant is composed of calcium cations ##Ca^(2=)## and sulfite anions ##SO_3^(2-)##. This means that you have
    ##color(blue)(CaSO_text(3(s]) -> ? + ?)##
    Next focus on the oxide. Since calcium cations have a ##2+## charge and oxygen anions have a ##2-## charge &calcium oxide will simply be
    ##Ca^(2+)O^(2-) implies Ca_2O_2 implies CaO##
    You now have
    ##color(blue)(CaSO_text(3(s]) -> CaO_text((s]) + ?)##
    Finally focus on sulfur dioxide. Sulfur dioxide is a which means that you can conclude from its name that it contains
    Therefore you can say that sulfur dioxide is ##SO_2##. This gives you
    ##color(blue)(CaSO_text(3(s]) -> CaO_text((s]) + SO_text(2(g]) uarr)##
    Since this equation is already balanced this will be your answer.

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