Balance each of the equations below, and identify the type of reaction in each case:

here are the balanced equations and their reaction types: a) Balance each of the equations below, and identify the type of reaction in each case: i) Al_4(CO_3)_3 → Al_2O_3 + CO Step 1: Balance Al atoms. There are 4 Al on the left, so place a coefficient of 2 in front of Al_2O_3. Al_4(CO_3)_3 → 2Al_2O_3 + CO Step 2: Balance C atoms. There are 3 C on the left, so place a coefficient of 3 in front of CO. Al_4(CO_3)_3 → 2Al_2O_3 + 3CO Step 3: Check O atoms. Left: 3 × 3 = 9. Right: (2 × 3) + (3 × 1) = 6 + 3 = 9. All atoms are balanced. Balanced Equation: Al_4(CO_3)_3 → 2Al_2O_3 + 3CO Type of Reaction: Decomposition reaction (one compound breaks down into simpler substances). ii) BaO_2 + H_2SO_4 → BaSO_4 + H_2O Assumption: Based on common chemical reactions, the product H_2O is likely a typo and should be H_2O_2 (hydrogen peroxide), as barium peroxide reacts with sulfuric acid to produce barium sulfate and hydrogen peroxide. The equation does not balance with H_2O as a product. Step 1: Assuming the product is H_2O_2, let's check the atoms. Reactants: Ba: 1, O: 2+4=6, H: 2, S: 1 Products: Ba: 1, O: 4+2=6, H: 2, S: 1 All atoms are already balanced. Balanced Equation: BaO_2 + H_2SO_4 → BaSO_4 + H_2O_2 Type of Reaction: Double displacement reaction (ions are exchanged between two compounds). iii) Ba(ClO_3)_2 → BaCl_2 + O_2 Step 1: Balance O atoms. There are 6 O on the left, so place a coefficient of 3 in front of O_2. Ba(ClO_3)_2 → BaCl_2 + 3O_2 Step 2: Check Ba and Cl atoms. Ba: 1 on both sides. Cl: 2 on both sides. All atoms are balanced. Balanced Equation: Ba(ClO_3)_2 → BaCl_2 + 3O_2 Type of Reaction: Decomposition reaction (one compound breaks down into simpler substances). iv) Cu + HNO_3 → Cu(NO_3)_2 + NO + H_2O Step 1: This is a redox reaction. Balance the oxidation and reduction changes. Cu goes from 0 to +2 (oxidized, change of +2). N goes from +5 in HNO_3 to +2 in NO (reduced, change of -3). To balance electrons, we need 3 Cu and 2 NO. Step 2: Start by placing coefficients for Cu and NO: 3Cu + HNO_3 → 3Cu(NO_3)_2 + 2NO + H_2O Step 3: Balance N atoms. On the right, there are 3 × 2 = 6 N in Cu(NO_3)_2 and 2 N in NO, totaling 8 N. So, place a coefficient of 8 in front of HNO_3. 3Cu + 8HNO_3 → 3Cu(NO_3)_2 + 2NO + H_2O Step 4: Balance H atoms. There are 8 H on the left. On the right, we need 8 H, so place a coefficient of 4 in front of H_2O. 3Cu + 8HNO_3 → 3Cu(NO_3)_2 + 2NO + 4H_2O Step 5: Check O atoms. Left: 8 × 3 = 24. Right: (3 × 2 × 3) + (2 × 1) + (4 × 1) = 18 + 2 + 4 = 24. All atoms are balanced. Balanced Equation: 3Cu + 8HNO_3 → 3Cu(NO_3)_2 + 2NO + 4H_2O Type of Reaction: Redox reaction (oxidation states of elements change). v) H_3BO_3 → H_3B_3O_6 + H_2O Step 1: Balance B atoms. There is 1 B on the left and 3 B on the right, so place a coefficient of 3 in front of H_3BO_3. 3H_3BO_3 → H_3B_3O_6 + H_2O Step 2: Balance H and O atoms. Reactants: H: 3 × 3 = 9, O: 3 × 3 = 9. Products: H: 3 (from H_3B_3O_6) + 2 (from H_2O) = 5. O: 6 (from H_3B_3O_6) + 1 (from H_2O) = 7. We need 9-3=6 more H and 9-6=3 more O on the product side. This corresponds to 3 molecules of H_2O. So, change the coefficient of H_2O to 3. 3H_3BO_3 → H_3B_3O_6 + 3H_2O Step 3: Check all atoms. Reactants: H: 9, B: 3, O: 9. Products: H: 3 + (3 × 2) = 9, B: 3, O: 6 + (3 × 1) = 9. All atoms are balanced. Balanced Equation: 3H_3BO_3 → H_3B_3O_6 + 3H_2O Type of Reaction: Decomposition reaction (specifically, a dehydration reaction where water is removed). vi) P_4 + P_2 + H_2O → PH_4I + H_3PO_4 Assumption: The reactant P_2 is likely a typo for I_2 (iodine), as P_2 is not a common stable reactant and the product PH_4I contains iodine. We will balance the reaction: P_4 + I_2 + H_2O → PH_4I + H_3PO_4. This is a disproportionation reaction of phosphorus. Step 1: Balance P, I, H, and O atoms using a system of equations or by inspection. Let the coefficients be aP_4 + bI_2 + cH_2O → dPH_4I + eH_3PO_4. P: 4a = d+e I: 2b = d H: 2c = 4d+3e O: c = 4e Step 2: From c=4e, substitute into the H equation: 2(4e) = 4d+3e 8e = 4d+3e 5e = 4d. Step 3: To find integer coefficients, let d=5k and e=4k. The smallest integers for d and e that satisfy 5e=4d are d=5 and e=4. Step 4: Substitute d=5 and e=4 into the other equations. From I: 2b = 5 b = 2.5. From P: 4a = 5+4 4a = 9 a = 2.25. Step 5: To get whole numbers, multiply all coefficients by 4 (the least common multiple of the denominators for a and b). a = 2.25 × 4 = 9 b = 2.5 × 4 = 10 d = 5 × 4 = 20 e = 4 × 4 = 16 Step 6: Calculate c using c=4e: c = 4 × 16 = 64. Balanced Equation: 9P_4 + 10I_2 + 64H_2O → 20PH_4I + 16H_3PO_4 Type of Reaction: Redox reaction (specifically, a disproportionation reaction where phosphorus is both oxidized and reduced). Got more? Send 'em.