How do you balance and translate this reaction: #Ca(OH)_2(aq) + H_2SO_4(aq) -> H_2O(l) + CaSO_4(s)#?

1 Answer
Mar 23, 2017

First balance the reaction

#Ca(OH)_2 + H_2SO_4 = H_2O + CaSO4#

Ca = 1#color (white)(mm)+color(white)(l)H =2 =color(white)()H=2 + color(white)(l)Ca = 1#
H = 2 #color (white)(mm)+color(white)(l)S=1 =color(white)(l)O = 1 + color(white)(l)S = 1#
O = 2 #color (white)(mm)+color(white)(l)O=4 = color(white)(MMMM)O = 1#
We find that 2atoms of hydrogen and 1 atom of O are unused.

If we combine these we can obtain 1mole of H2O

#Ca(OH)_2 + H_2SO_4 = 2H_2O + CaSO4#

Now you can write down the ionic reaction

#Ca^(2+) + 2OH^(-) + 2H^(+) + SO4^(2-) = 2H_2O+ Ca^(2+) + SO4^(2-)#

(I know that SO4^-2 is confusing. So I recommend you that to determine the charge of an ion . Use the zero sum rule .)

The zero sum rule says that the addition of all charges in molecule is 0.

For example

If you know the charge of Ca in CaSO4 you can determine the charge of SO4

the equation would be like this

0 = 2+ (charge of calcium ion) + x

Solve for x and you would get the charge of SO4.

So ionic reaction=
#Ca^(2+)+2OH^(-)+2H^(+)+SO4^(2-)=2H_2O+Ca^(2+)+SO4^(2-)#

Cut out the same ions on both sides

#cancel(Ca^(2+))+2OH^(-)+2H^+ + cancel(SO4^(2-)) = 2H_2O + cancel(Ca^+2) + cancel(SO4^-2)#

Ionic reaction

#2OH^-) + 2H^+ = 2H_2O#

= #OH^-) + H^+ = H_2O#