Question

How much of 30% HCl and 45% HCl would be needed to mix and produce 50 liters of 35% HCl?

Answer 1

You have to make an equation .3x+.45y=.35(50) as well as the equation x+b=50. (x is the volume of the 30% solution and b is the y is the volume of the 45% solution). You can make x=50-y and substitute that in for x in the first equation to get the equation: .3(50-y)+.45y=.35(50). You have to solve for y. 15-.3y+.45y=17.5 .15y=2.5 Y=16.67 Then solve for x using x=50-y to get x=33.33. Therefore you need 33.33L of 30% solution and 16.67L of 45% solution. The reason why this method works is that when you are given a percent concentration you are getting the ratio showing how much solute there is in the solution and if you multiply the volume of solution by the percent solute, you are finding the amount of solute. Using that, you can think of the first equation being amount HCl from one solution plus amount HCl from the other solution equals the amount HCl in the final solution. I hope this helps let me know in the comments if anything is unclear.

Answer 2

42.8572 L of 30% HCl and 7.1428 L of 45% HCl would be needed to mix and produce 50 liters of 35% HCl.

Step-by-step explanation:

Let volume of 30% HCl solution required to make 35% of HCl solution be x

Volume of 45% of HCl required to make 35% of HCl solution be y

Volume of solution required = 50 L

x + y = 50 ..(1)

30% of x +45 % of y = 35% of 50

`(30)/(100)* x+(45)/(100)* y=(35)/(100)* 50`

3x + 45y = 1750 ..(2)

Solving equation (1) and (2):

x = 42.8572 L

y =7.1428 L

42.8572 L of 30% HCl and 7.1428 L of 45% HCl would be needed to mix and produce 50 liters of 35% HCl.