Problem strategy: Look at the chemical equation for the elements that occurs in only one substance on each side of the equation.Begin by balancing the equation in one of these atoms.Use the number of atoms on the left side of the arrow as the coefficient of the substance containing that element on the right side, and vice versa.After balancing one element in an equation , the rest become easier.
Part (a) H3PO4 ----------> H3PO4 + PH3
oxygen occurs in just one of the products H3PO4. It is therefore easiest to balance O atoms first.To do this note that H3PO4 has three O atoms; use 3 as coefficient of H3PO4 on the right side.There ar four O atoms in H3PO4 on the right side; use 4 as the coefficent of H3PO4 on the left side.
4H3PO4 ----------> 3H3PO4 + PH3
this equation is now also balanced in P and H atoms.
(B) Ca + H20 -------> Ca(OH)2 + H2
The equation is balanced in Ca as it stands
1Ca + H20 -------> 1Ca(OH)2 + H2
O occurs in only one reactant and in only one product, so they are balanced next.
Ca + 2H20 -------> Ca(OH)2 + H2
The equation is now also balanced in H atoms.
(c) To balance the equation in Fe atoms, you write
1Fe2(SO4)3 + NH3 + H20 ----------> 1Fe(OH)3 + (NH4)2SO4
You balanced the S atoms by placing the coefficent 3 for (NH4)So4.
1Fe2(SO4)3 + NH3 + H20 ----------> 2Fe(OH)3 + 3(NH4)2SO4
Now balance the N atoms
Fe2(SO4)3 + 6NH3 + H20 ----------> 2Fe(OH)3 + 3(NH4)2SO4
To balance the O atoms, you first count the number of O atoms on the right(18).Then you count the number of O atoms in substances on the left with known coefficients.There are 12 O's in Fe2(SO4)2; hence the number of remaining O's (in H20) must be 18-12 = 6.
Fe2(SO4)3 + 6NH3 + 6H20 ----------> 2Fe(OH)3 + 3(NH4)2SO4
this equation is now balanced in H atoms.