II. Balancing Equations Ch. 7/8 – Chemical Reactions II. Balancing Equations C. Johannesson

Coefficient  subscript = # of atoms A. Balancing Steps 1. Write the unbalanced equation. 2. Count atoms on each side. 3. Add coefficients to make #s equal. Coefficient  subscript = # of atoms 4. Reduce coefficients to lowest possible ratio, if necessary. 5. Double check atom balance!!! C. Johannesson

B. Helpful Tips Balance one element at a time. Update ALL atom counts after adding a coefficient. If an element appears more than once per side, balance it last. Balance polyatomic ions as single units. “1 SO4” instead of “1 S” and “4 O” C. Johannesson

C. Balancing Example 2 Al + CuCl2  Cu + AlCl3 Al Cu Cl 3 3 2 2  1 1 Aluminum and copper(II) chloride react to form copper and aluminum chloride. 2 Al + CuCl2  Cu + AlCl3 Al Cu Cl 3 3 2 2  1 1 2 3  2  6 3  6   3 C. Johannesson

Mr. Rapps slides next C. Johannesson

Compare the numbers of each kind of atom in the balanced equation with the numbers of each kind of atom in the sketched representation. Both the equation and the sketch have the same number of atoms in the reactants and in the products.

Practice Al(s) + O2(g) ---> Al2O3(s) Inventory: 1Al and 2O’s on left and 2Al’s and 3O’s on right. Start with Al Need 2 on left to balance right 2Al(s) + O2(g) ---> Al2O3(s) Al is balanced but O is not. Double everything but O2

Practice 4Al(s) + O2(g) ---> 2Al2O3(s) Al is still balanced, now we have 2O’s on left and 6O’s on right. How can we balance that? 3x2 = 6 so we need 3 O2 molecules 4Al(s) + 3O2(g) ---> 2Al2O3(s) Inventory: 4Al’s and 6O’s on left and 4Al’s and 6O’s on right It’s balanced

Practice Balance: FeCl3 + NaOH  Fe(OH)3 + NaCl Inventory: 1Fe, 3Cl’s, 1Na and 1OH on left and 1Fe, 3OH’s, 1Na and 1Cl on right Balance Cl first  need 3 on right FeCl3 + NaOH  Fe(OH)3 + 3NaCl Na is now unbalanced  need 3 on left now FeCl3 + 3NaOH  Fe(OH)3 + 3NaCl Inventory again: 1Fe, 3Cl, 3Na, and 3OH on left and 1Fe, 3OH, 3Na, and 3Cl on right It’s balanced!!!

Practice Ba(NO3)2 + Al2(SO4)3  BaSO4 + Al(NO3)3 Inventory: 1Ba, 2NO3’s, 2Al’s and 3SO4’s on left and 1Ba, 1SO4, 1Al, and 3NO3’s on right Balance SO4’s first  need 3 on right Ba(NO3)2 + Al2(SO4)3  3BaSO4 + Al(NO3)3 Balance Ba next  need 3 on left 3Ba(NO3)2 + Al2(SO4)3  3BaSO4 + Al(NO3)3 Balance NO3’s next  6 on left and 3 on right, use 2 as coefficient

Practice 3Ba(NO3)2 + Al2(SO4)3  3BaSO4 + 2Al(NO3)3 That gives 6NO3’s on both sides Inventory: 3Ba’s, 6NO3’s, 2Al’s and 3SO4’s on left and 3Ba’s, 3SO4’s, 2Al’s, and 6NO3’s on right. It’s balanced

Practice Ca(OH)2 + H3PO4  Ca3(PO4)2 + H2O Inventory: 1Ca, 5H’s, 6O’s and 1P on left and 3Ca’s, 9O’s, 2H’s, and 2 P on right. Balance Ca first  need 3 on left 3Ca(OH)2 + H3PO4  Ca3(PO4)2 + H2O Balance PO4’s next  need 2 on left 3Ca(OH)2 + 2H3PO4  Ca3(PO4)2 + H2O Inventory again for H’s and O’s: 6O’s+8O’s(14) and 6H’s+6H’s(12) on left and 8O’s+1O’s(9) and 2H’s on right

Practice To get 12H’s on right we can use 6 as a coefficient for water 3Ca(OH)2 + 2H3PO4  Ca3(PO4)2 + 6H2O Inventory: 3Ca’s, 2P’s, (6+6=12)H’s, and (6+8=14)O’s on left and 3Ca’s, 2P’s, (6x2=12)H’s and (8+6=14)O’s on right It’s balanced

Practice ____C3H8(g) + _____ O2(g) ----> _____CO2(g) + _____ H2O(g) ____B4H10(g) + _____ O2(g) ---- > ___ B2O3(g) + _____ H2O(g) 5 3 4 2 11 4 10

Practice NH4NO3 ---> N2O + 2 H2O Carbon monoxide reacts with nitrogen monoxide to form carbon dioxide and nitrogen gas (pairs) Equation: CO + NO  CO2 + N2 CO + NO  CO2 + N2 2 2 2

Practice Phosphorus trichloride reacts with water to form hydrochloric acid and phosphorous acid. Equation: PCl3 + H2O  HCl + H3PO3 Magnesium nitride reacts with water to form magnesium hydroxide and ammonia Mg3N2 + H2O  Mg(OH)2 + NH3 PCl3 + H2O  HCl + H3PO3 3 3 Mg3N2 + H2O  Mg(OH)2 + NH3 6 3 2