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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Step 1: Balance Carbon (C) atoms. There are 3 carbon atoms on the reactant side in C_3H_5N_3O_9. To balance this, place a coefficient of 3 in front of CO_2 on the product side. C_3H_5N_3O_9 → 3CO_2 + H_2O + N_2 + O_2 Step 2: Balance Hydrogen (H) atoms. There are 5 hydrogen atoms on the reactant side in C_3H_5N_3O_9. To balance this, place a coefficient of (5)/(2) in front of H_2O on the product side. C_3H_5N_3O_9 → 3CO_2 + (5)/(2)H_2O + N_2 + O_2 Step 3: Balance Nitrogen (N) atoms. There are 3 nitrogen atoms on the reactant side in C_3H_5N_3O_9. To balance this, place a coefficient of (3)/(2) in front of N_2 on the product side. C_3H_5N_3O_9 → 3CO_2 + (5)/(2)H_2O + (3)/(2)N_2 + O_2 Step 4: Balance Oxygen (O) atoms. Count oxygen atoms on both sides: Reactant side: 9 O atoms in C_3H_5N_3O_9. Product side: (3 × 2) O in 3CO_2 + ((5)/(2) × 1) O in (5)/(2)H_2O + (x × 2) O in xO_2. So, 9 = 6 + (5)/(2) + 2x. 9 = (12)/(2) + (5)/(2) + 2x 9 = (17)/(2) + 2x 18 = 17 + 4x (multiplying by 2 to clear the fraction) 1 = 4x x = (1)/(4) Place a coefficient of (1)/(4) in front of O_2. C_3H_5N_3O_9 → 3CO_2 + (5)/(2)H_2O + (3)/(2)N_2 + (1)/(4)O_2 Step 5: Clear fractions by multiplying all coefficients by the least common multiple (LCM) of the denominators (2, 2, 4), which is 4. 4(C_3H_5N_3O_9) → 4(3CO_2) + 4((5)/(2)H_2O) + 4((3)/(2)N_2) + 4((1)/(4)O_2) 4C_3H_5N_3O_9 → 12CO_2 + 10H_2O + 6N_2 + O_2 Step 6: Verify the balance of all atoms. Reactants: C = 12, H = 20, N = 12, O = 36 Products: C = 12, H = 20, N = 12, O = (12 × 2) + (10 × 1) + (1 × 2) = 24 + 10 + 2 = 36 All atoms are balanced. The balanced chemical equation is: 4C_3H_5N_3O_9 → 12CO_2 + 10H_2O + 6N_2 + O_2 What's next? Send 'em!