# Chemical solution question.

• January 29th 2011, 09:15 PM
grooverandshaker
Chemical solution question.
A chemical manufacturer has an order for 500 litres of a 25% acid solution (i.e. 25% by volume is acid). Solutions of 30% and 18% are available in stock.
How much acid is required to produce 500 litres of 25% acid solution?
The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
Let x denote the amount of 30% solution required.
Let y denote the amount of 18% solution required.
Use simultaneous equations in x and y to determine the amount of each solution.
I'm pretty sure once I find what the simultaneous equations are, I will be able to solve them.
Thanks. :D
• January 29th 2011, 09:29 PM
earboth
Quote:

Originally Posted by grooverandshaker
A chemical manufacturer has an order for 500 litres of a 25% acid solution (i.e. 25% by volume is acid). Solutions of 30% and 18% are available in stock.
How much acid is required to produce 500 litres of 25% acid solution?
The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
Let x denote the amount of 30% solution required.
Let y denote the amount of 18% solution required.
Use simultaneous equations in x and y to determine the amount of each solution.
I'm pretty sure once I find what the simultaneous equations are, I will be able to solve them.
Thanks. :D

1. The 1st equation refers to the volumes of the liquids:

$x+y=500$

2. The 2nd equation refers to the volume of the acid:

$0.3 \cdot x + 0.18 \cdot y = 0.25 \cdot 500$

3. Solve for x and y.
• January 30th 2011, 04:12 PM
bjhopper
mixture problem
an easier to way to solve this

draw a concentration line

18 25 30
lean mix rich
7 5

5 parts lean + 7 parts rich = 12 parts mix

208.3 liter + 291.6 liter = 499.9liters mix

bjh
• January 30th 2011, 07:03 PM
grooverandshaker
Thank you for all of your help! =)