A chemist mixes a 20% acid solution with a 50% acid solution to obtain 30 liters of a 35% acid solution. How many liters of the 20% solution are used?
["Title: Solving a Classic Acid Mixing Problem: How Many Liters of 20% Solution Are Used?", "Meta Description:\nLearn how to solve the classic chemistry mixture problem: mixing a 20% acid solution with a 50% acid solution to produce 30 liters of a 35% acid solution. Discover the step-by-step calculation and find out exactly how many liters of the 20% solution are needed.", "---", "### Solution to the Acid Mixing Problem", "Imagine a chemist preparing a 30-liter mixture of acid solutions to create a precise 35% concentration. To achieve this, they combine two acid solutions: a 20% acid solution and a 50% acid solution. The key question is: how many liters of the 20% solution are required?", "#### Step 1: Define Variables\nLet:\n- ( x ) = liters of the 20% acid solution\n- ( 30 - x ) = liters of the 50% acid solution (since total volume is 30 liters)", "#### Step 2: Set Up the Concentration Equation\nThe total amount of pure acid in the final mixture must equal the acid contributed by each component:", "Acid from 20% solution: ( 0.20x )\nAcid from 50% solution: ( 0.50(30 - x) )\nTotal acid in final 35% solution: ( 0.35 \ imes 30 = 10.5 ) liters", "Now write the equation:\n[\n0.20x + 0.50(30 - x) = 10.5\n]", "#### Step 3: Solve the Equation\nExpand the left side:\n[\n0.20x + 15 - 0.50x = 10.5\n]", "Combine like terms:\n[\n-0.30x + 15 = 10.5\n]", "Subtract 15 from both sides:\n[\n-0.30x = -4.5\n]", "Divide by -0.30:\n[\nx = \frac{-4.5}{-0.30} = 15\n]", "#### Step 4: Conclusion\nThe chemist uses 15 liters of the 20% acid solution.", "Then, the remaining volume is ( 30 - 15 = 15 ) liters of the 50% solution.", "#### Final Verification\nTotal acid:\n( 0.20 \ imes 15 = 3 ) liters from 20% solution\n( 0.50 \ imes 15 = 7.5 ) liters from 50% solution\nTotal acid = ( 3 + 7.5 = 10.5 ) liters\nConcentration = ( \frac{10.5}{30} = 0.35 = 35% ), which matches the target.", "---", "### Why This Matters", "This problem exemplifies a classic application of alkali-denitrate blending or acid-base chemistry, used in labs and industry to prepare precise chemical mixtures. Understanding how to set up and solve these equations helps chemists and chemical engineers optimize processes for accuracy and safety.", "Whether you're preparing reagents for analysis or industrial applications, mastering mixture problems ensures reliable results every time.", "---", "Keywords: acid solution mixture, chemistry problem, 20% acid, 50% acid, mixing solutions, chemical concentration calculation, how much 20% acid to use, acid solutions, solution preparation.", "Tools: This calculation uses algebra and concentration formulas, often applied in stoichiometry and laboratory preparation. For quick reference, use linear equation models or mixing equation solvers to verify results.", "---", "Stay accurate, stay safe — understanding how to mix chemicals correctly is essential for successful science and engineering!"]
