# Help me on mixture problem T_T

• Sep 2nd 2006, 05:35 AM
Help me on mixture problem T_T
A chemical engineer mixed 40ml of 80% HCl solution w/ 60ml of 12% HCL solution. He used a portion of the resulting mixtures & replace it with distilled water / H2O. If the new solution is 5.2% HCl, Howmuch of the original mixture did he used?
• Sep 2nd 2006, 06:15 AM
topsquark
Quote:

Originally Posted by ^_^Engineer_Adam^_^
A chemical engineer mixed 40ml of 80% HCl solution w/ 60ml of 12% HCL solution. He used a portion of the resulting mixtures & replace it with distilled water / H2O. If the new solution is 5.2% HCl, Howmuch of the original mixture did he used?

The new solution is 5.2% HCl before or after the distilled water was added? If it is after we need to know how much distilled water was used.

-Dan
• Sep 3rd 2006, 03:27 AM
Random333
Quote:

Originally Posted by ^_^Engineer_Adam^_^
A chemical engineer mixed 40ml of 80% HCl solution w/ 60ml of 12% HCL solution. He used a portion of the resulting mixtures & replace it with distilled water / H2O. If the new solution is 5.2% HCl, Howmuch of the original mixture did he used?

Well we have 40mL+60mL, so the total solution is 100mL.
And have 32mL and 7.2mL of HCL respectively giving a total of 39.2mL of HCL.

We want to reduce this amount of HCL to 5.2mL if we replace all the solution used with the exact same amount of distilled water.

Therefore amount of solution to be used:
$\frac{34}{39.2} \times 100$