How do you balance #Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2#?

1 Answer
Meave60
Apr 24, 2018

#"Ca"_3("PO"_4)_2+4"H"_3"PO"_4"##rarr##3"Ca(H"_2"PO"_4)_2"#

Explanation:

Given equation:

#"Ca"_3("PO"_4)_2+"H"_3"PO"_4"##rarr##"Ca(H"_2"PO"_4)_2"#

There are three Ca atoms on the left-hand side (LHS), and one on the right-hand side (RHS). Place a coefficient of #3# in front of #"Ca(H"_2"PO"_4)_2#.

#"Ca"_3("PO"_4)_2+"H"_3"PO"_4"##rarr##color(teal)3"Ca(H"_2"PO"_4)_2"#

Now there are twelve H atoms on the RHS #(3xx2xx2)#, and three on the LHS. Place a coefficient of #4# in front of #"H"_3"PO"_4"#.

#"Ca"_3("PO"_4)_2+color(magenta)4"H"_3"PO"_4"##rarr##color(teal)3"Ca(H"_2"PO"_4)_2"#

Now lets count the number of atoms of each element on each side of the equation.

#"LHS:"# #"3 Ca atoms"#, #"6 P atoms"#, #"12 H atoms"#, #"24 O atoms"#

#"RHS:"# #"3 Ca atoms"#, #"6 P atoms"#, #"12 H atoms"#, #"24 O atoms"#

Since there are the same number of atoms of each element on both sides of the equation, it is balanced.

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