This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions for questions 7 and 8. 7. Balance the equations below. a) Al + Cl_2 Al_2Cl_6 Step 1: Balance Al atoms. There are 2 Al on the right, so put a coefficient of 2 in front of Al on the left. 2Al + Cl_2 Al_2Cl_6 Step 2: Balance Cl atoms. There are 6 Cl on the right, and 2 Cl on the left. Put a coefficient of 3 in front of Cl_2 on the left. 2Al + 3Cl_2 Al_2Cl_6 The balanced equation is: 2Al + 3Cl_2 Al_2Cl_6 b) C_2H_6 + O_2 CO_2 + H_2O Step 1: Balance C atoms. There are 2 C on the left, so put a coefficient of 2 in front of CO_2 on the right. C_2H_6 + O_2 2CO_2 + H_2O Step 2: Balance H atoms. There are 6 H on the left, so put a coefficient of 3 in front of H_2O on the right. C_2H_6 + O_2 2CO_2 + 3H_2O Step 3: Balance O atoms. There are 2 × 2 + 3 × 1 = 7 O atoms on the right. There are 2 O atoms on the left. To get 7 O atoms on the left, use a coefficient of (7)/(2) for O_2. C_2H_6 + (7)/(2)O_2 2CO_2 + 3H_2O Step 4: Multiply the entire equation by 2 to remove the fraction. 2C_2H_6 + 7O_2 4CO_2 + 6H_2O The balanced equation is: 2C_2H_6 + 7O_2 4CO_2 + 6H_2O c) TiCl_4 + Na Ti + NaCl Step 1: Balance Cl atoms. There are 4 Cl on the left, so put a coefficient of 4 in front of NaCl on the right. TiCl_4 + Na Ti + 4NaCl Step 2: Balance Na atoms. There are 4 Na on the right, so put a coefficient of 4 in front of Na on the left. TiCl_4 + 4Na Ti + 4NaCl The balanced equation is: TiCl_4 + 4Na Ti + 4NaCl d) H_2O_2 H_2O + O_2 Step 1: Balance O atoms. There are 2 O on the left and 3 O on the right. To balance, try to make the oxygen atoms on the right an even number by putting a coefficient of 2 in front of H_2O. H_2O_2 2H_2O + O_2 Step 2: Now there are 2 × 1 + 2 = 4 O atoms on the right and 2 O atoms on the left. Put a coefficient of 2 in front of H_2O_2 on the left. 2H_2O_2 2H_2O + O_2 Step 3: Check H atoms. There are 2 × 2 = 4 H atoms on the left and 2 × 2 = 4 H atoms on the right. The equation is balanced. The balanced equation is: 2H_2O_2 2H_2O + O_2 8. Complete the following | Compound | Element | | :------------------- | :----------- | | 1. Water | Hydrogen | | 2. Common Salt | Sodium | | 3. Hydrochloric acid | Hydrogen | | 4. Carbon (IV) oxide | Carbon | | 5. Ammonium chloride | Nitrogen | | 6. Sulphuric (VI) acid | Sulfur | | 7. Nitrogen (IV) oxide | Nitrogen | That's 2 down. 3 left today — send the next one.