Summary
Highlights
The video begins with balancing the reaction of zinc and nitrogen to produce zinc nitride. Zinc has a +2 charge, nitrogen is diatomic (N2), and nitride has a -3 charge. The process involves writing the skeleton equation Zn + N2 → Zn3N2, and then balancing by adding coefficients, resulting in 3Zn + N2 → Zn3N2.
This section explains balancing the reaction between cobalt(III) bromide and fluorine to yield cobalt(III) fluoride and bromine. Cobalt(III) has a +3 charge, bromide is -1, and fluorine is diatomic. The equation is CoBr3 + F2 → CoF3 + Br2. The instructor demonstrates balancing the elements systematically, arriving at 2CoBr3 + 3F2 → 2CoF3 + 3Br2.
The third example covers the decomposition of potassium hydroxide into potassium oxide and water. Potassium has a +1 charge, hydroxide is -1, and oxide is -2. The balanced equation derived is 2KOH → K2O + H2O.
This part details the reaction of lithium with tin(IV) sulfite, forming lithium sulfite and tin. Tin(IV) has a +4 charge, sulfite (SO3) has -2, and lithium has +1. Polyatomic ions (sulfite) are identified and balanced first. The final balanced equation shown is 4Li + Sn(SO3)2 → 2Li2SO3 + Sn.
The fifth example is a double displacement reaction between silver carbonate and iron(II) acetate, producing iron(II) carbonate and silver acetate. Charges are assigned: Silver (+1), Carbonate (-2), Iron(II) (+2), Acetate (-1). Polyatomic ions are kept together. The balanced equation is Ag2CO3 + Fe(C2H3O2)2 → FeCO3 + 2AgC2H3O2.
This section addresses balancing the reaction of copper(I) phosphate and nickel, which produces nickel(II) phosphate and copper. Copper(I) is +1, phosphate (PO4) is -3, and nickel is a free element. Balancing polyatomic ions first is key. The result is 2Cu3PO4 + 3Ni → Ni3(PO4)2 + 6Cu.
This part focuses on balancing the reaction between sodium iodide and magnesium chlorate. Sodium is +1, iodide is -1, magnesium is +2, and chlorate (ClO3) is -1. The resulting balanced equation is 2NaI + Mg(ClO3)2 → MgI2 + 2NaClO3.
The video then covers the reaction of iron(III) chromate and lithium nitrate. Iron(III) is +3, chromate (CrO4) is -2, lithium is +1, and nitrate (NO3) is -1. This complex double displacement reaction requires careful balancing of multiple polyatomic ions and elements. The final balanced equation is Fe2(CrO4)3 + 6LiNO3 → 2Fe(NO3)3 + 3Li2CrO4.
This example tackles the decomposition of aluminum chlorate (Al(ClO3)3) into aluminum chloride (AlCl3) and oxygen (O2). Since the polyatomic ion (chlorate) breaks apart, each element is listed separately. The instructor demonstrates using a fraction (9/2) for oxygen and then multiplying the entire equation by 2 to clear the fraction, resulting in 2Al(ClO3)3 → 2AlCl3 + 9O2.
The final problem involves the reaction of barium with water, producing barium hydroxide and hydrogen gas. Barium is +2, water is H2O, hydroxide is -1, and hydrogen is diatomic. The balanced equation is Ba + 2H2O → Ba(OH)2 + H2.