Question 2.12: Balancing Simple Equations Balance the following equations. ...
Balancing Simple Equations
Balance the following equations.
a. H_3PO_3 → H_3PO_4 + PH_3\\ b. Ca + H_2O → Ca(OH)_2 + H_2\\ c. Fe_2(SO_4)_3 + NH_3 + H_2O → Fe(OH)_3 + (NH_4)_2SO_4PROBLEM STRATEGY
Look at the chemical equation for atoms of elements that occur 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.
a. Oxygen occurs in just one of the products (H_3PO_4). It is therefore easiest to balance O atoms first. To do this, note that H_3PO_3 has three O atoms; use 3 as the coefficient of H_3PO_4 on the right side. There are four O atoms in H_3PO_4 on the right side; use 4 as the coefficient of H_3PO_3 on the left side.
\underline{4}H_3PO_3 → \underline{3}H_3PO_4 + PH_3This equation is now also balanced in P and H atoms. Thus, the balanced equation is
4H_3PO_3 → 3H_3PO_4 + PH_3b. The equation is balanced in Ca atoms as it stands.
\underline{1}Ca + H_2O → \underline{1}Ca(OH)_2 + H_2O atoms occur in only one reactant and in only one product, so they are balanced next.
\underline{1}Ca + \underline{2}H_2O → \underline{1}Ca(OH)_2 + H_2The equation is now also balanced in H atoms. Thus, the answer is
Ca + 2H_2O → Ca(OH)_2 + H_2c. To balance the equation in Fe atoms, you write
\underline{1}Fe_2(SO_4)_3 + NH_3 + H_2O → \underline{2}Fe(OH)_3 + (NH_4)_2SO_4You balance the S atoms by placing the coefficient 3 for (NH_4)_2SO_4
\underline{1}Fe_2(SO_4)_3 + NH_3 + H_2O → \underline{2}Fe(OH)_3 + \underline{3}(NH_4)_2SO_4Now you balance the N atoms.
\underline{1}Fe_2(SO_4)_3 + \underline{6}NH_3 + H_2O → \underline{2}Fe(OH)_3 + \underline{3}(NH_4)_2SO_4To 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 Fe_2(SO_4)_3; hence, the number of remaining O’s (in H_2O) must be 18 – 12 = 6.
\underline{1}Fe_2(SO_4)_3 + \underline{6}NH_3 + \underline{6}H_2O → \underline{2}Fe(OH)_3 + \underline{3}(NH_4)_2SO_4Finally, note that the equation is now balanced in H atoms. The answer is
Fe_2(SO_4)_3 + 6NH_3 + 6H_2O → 2Fe(OH)_3 + 3(NH_4)_2SO_4Check each final equation by counting the number of atoms of each element on both sides of the equations.