Respuesta :
Let x be the volume (liters) of pure (at 100%) acid needed
Let y be the volume (liters) of the other acid (at 10%) needed
The final solution will be:
a) x+ y = 63 liters, AND their respective concentration in acid: is
100% x & 10% :y that will generate 63 liters at 20%
b) x +0.1 y = 63x 0.2 = 12.6
Let's solve this system of 2 equations:
x + y =63
x + 0.1 y = 12.6
Solving it will give you:
x= 7 liters at 100%
y = 56 liters at 10%
Let y be the volume (liters) of the other acid (at 10%) needed
The final solution will be:
a) x+ y = 63 liters, AND their respective concentration in acid: is
100% x & 10% :y that will generate 63 liters at 20%
b) x +0.1 y = 63x 0.2 = 12.6
Let's solve this system of 2 equations:
x + y =63
x + 0.1 y = 12.6
Solving it will give you:
x= 7 liters at 100%
y = 56 liters at 10%
(63-x)= amount of 10% solution wanted
0.10(63-x) + x =0.20(63) =
6.3-0.1x+x=12.6
0.9x=6.3
X=7
63-7 = 56 liters
Check:
0.10(56) +7 = 0.2(63)
12.6 = 12.6