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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
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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 = 12:1


You can write two equations:

W = volume of water

H = 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 = 12H


Your final two equations are:

W + H = 55

W = 12H


Solve for W and H using substitution:

Substitute 12H for W in the first equation, and then solve for H

W + H = 55

12H + H = 55


Combine like terms

12H + H = 55

13H = 55


Divide each side of the equation by 13 like terms

13H = 55

13H / 13 = 55 / 13

H * (13 / 13) = 55 / 13

H * (1) = 55 / 13

H = 55 / 13

H = 4.2307692307692


H (volume of sodium hydroxide) = 4.23 gallons



Substitute 4.23 gallons for H in the first equation, and then solve for W

W + H = 55

W + 4.23 = 55


Subtract 4.23 from each side of the equation

W + 4.23 = 55

W + 4.23 - 4.23 = 55 - 4.23

W + 0 = 55 - 4.23

W = 55 - 4.23

W (volume of water) = 50.77 gallons


Compute 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



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