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How do you balance a single replacement reaction?

    You balance the equation for a single replacement reaction the same way as any other equation. You place coefficients in front of the chemical formulas so that the numbers of atoms of the are the same on both sides.
    Example 1.
    ##”Na”(“s”)## + ##”ZnCl”_2(“aq”)”## ##rarr”## ##”NaCl”(“aq”)## + ##”Zn”(“s”)##
    First we notice that there are two Cl atoms on the left side of the equation and one on the right, so we add a coefficient of 2 in front of ##”NaCl”##.
    ##”Na”(“s”)## + ##”ZnCl”_2(“aq”)”## ##rarr”## ##”2NaCl”(“aq”)## + ##”Zn”(“s”)##
    Now we have two Na atoms on the right side, but only one on the left side, so we add a coefficient of 2 in front of the Na.
    ##”2Na”(“s”)## + ##”ZnCl”_2(“aq”)”## ##rarr”## ##”2NaCl”(“aq”)## + ##”Zn”(“s”)##
    Now the equation is balanced. There are equal numbers of each element on both sides of the equation.
    If the equation involves a polyatomic ion, it is easier to balance the equation if you consider the polyatomic ion as a single unit.
    ##”Al”(“s”)## + ##”CuSO”_4″(“aq”)## ##rarr## ##”Al”_2″(SO”_4)_3″(“aq”)## + ##”Cu”(“s”)##
    If you consider the polyatomic sulfate ion ##”SO”_4^(2-)##. (Don’t worry about the charge. The chemical formulas for the containing the sulfate ion are correct.) If you look on the left side of the equation, there is one sulfate ion, and on the right side there are three sulfate ions. So place a coefficient of 3 in front of the ##”CuSO”_4″(“aq”)##.
    ##”Al”(“s”)## + ##”3CuSO”_4″(“aq”)## ##rarr## ##”Al”_2″(SO”_4)_3″(aq)## + ##”Cu”(“s”)##
    Now the Al needs to be balanced. There are two Al atoms on the right side and one on the left. Place a coefficient of 2 in front of the Al on the left side.
    ##”2Al”(“s”)## + ##”3CuSO”_4″(“aq”)## ##rarr## ##”Al”_2″(SO”_4)_3″(aq)## + ##”Cu”(“s”)##
    Now we need to balance the Cu. There are three Cu atoms on the left side and one on the right. So we need to place a coefficient of 3 in front of the Cu on the right side.
    ##”2Al”(“s”)## + ##”3CuSO”_4″(“aq”)## ##rarr## ##”Al”_2″(SO”_4)_3″(aq)## + ##”3Cu”(“s”)##
    Now the equation is balanced. There are equal numbers of atoms of each element on both sides, and equal numbers of on both sides.

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