Help with % of Constituents in a 55-gallon drum

by Kim
(San Diego, CA)

Mixture Problem

• Determine what are the percentages of water and sodium hydroxide are in a drum.

If I have a 55 gallon drum of water and sodium hydroxide where the mixture of water to Sodium Hydroxide is 12:1, what are the percentages of water and sodium hydroxide in the drum?

Comments for Help with % of Constituents in a 55-gallon drum

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 Aug 21, 2012 Rating Mixture Problem by: Staff Answer:The problem statement gives you two facts about the mixture:    1. Volume of water + volume of sodium hydroxide = 55 gal    2. Ratio of the volume of water to the volume of sodium hydroxide = 12:1You can write two equations:W = volume of waterH = volume of sodium hydroxide    1. Volume of water + volume of sodium hydroxide = 55 gal          W + H = 55    2. Ratio of the volume of water to the volume of sodium hydroxide = 12:1          W:H = 12:1          W/H = 12/1          W * (H / H) = (12/1) * H          W * (1) = (12/1) * H          W = (12/1) * H          W = 12 * H          W = 12HYour final two equations are:W + H = 55W = 12HSolve for W and H using substitution:Substitute 12H for W in the first equation, and then solve for HW + H = 5512H + H = 55Combine like terms12H + H = 5513H = 55Divide each side of the equation by 13 like terms13H = 5513H / 13 = 55 / 13H * (13 / 13) = 55 / 13H * (1) = 55 / 13H = 55 / 13H = 4.2307692307692H (volume of sodium hydroxide) = 4.23 gallonsSubstitute 4.23 gallons for H in the first equation, and then solve for WW + H = 55W + 4.23 = 55Subtract 4.23 from each side of the equationW + 4.23 = 55W + 4.23 - 4.23 = 55 - 4.23W + 0 = 55 - 4.23W = 55 - 4.23W (volume of water) = 50.77 gallonsCompute the percentage by volume% of H (% volume of sodium hydroxide) = (4.23 gallons / 55 gallons) * 100% of H (% volume of sodium hydroxide) = (4.23 / 55) * (gallons / gallons) * 100% of H (% volume of sodium hydroxide) = (4.23 / 55) * (1) * 100% of H (% volume of sodium hydroxide) = (4.23 / 55) * 100% of H (% volume of sodium hydroxide) = (0.0769090909091) * 100% of H (% volume of sodium hydroxide) = 7.69090909091% of H (% volume of sodium hydroxide) = 7.69 %% of W (% volume of volume of water) = (50.77 gallons / 55 gallons) * 100% of W (% volume of volume of water) = (50.77 / 55) * (gallons / gallons) * 100% of W (% volume of volume of water) = (50.77 / 55) * (1) * 100% of W (% volume of volume of water) = (50.77 / 55) * 100% of W (% volume of volume of water) = (0.9230909090909) * 100% of W (% volume of volume of water) = 92.30909090909% of W (% volume of volume of water) = 92.31 %>>> the final answer is:% of H (% volume of sodium hydroxide) = 7.69 %% of W (% volume of volume of water) = 92.31 %Thanks for writing. Staff www.solving-math-problems.com