How many grams do you need to take a 20% solution and 40% sulfuric acid solution to get 500 g

How many grams do you need to take a 20% solution and 40% sulfuric acid solution to get 500 g of a 30% sulfuric acid solution?

Let us find the mass of the solute m (r.v.) and the mass of water m (c) in the final solution:

m (r.v.) = m (solution) * w / 100%, where m (solution) is the mass of the solution, w is the mass fraction of the solute.

m (r.v.) = 500 * 30/100% = 150 g

m (in) = m (solution) – m (r.v.) = 500 – 150 = 350 g

The masses of solutes in the initial solutions will be calculated in the same way:

m (r.v. 20%) = m (r.v. 20%) * w1 / 100% and m (r.v. 40%) = m (r.v. 40%) * w2 / 100%,

where w1 and w2 are the mass fractions of the solute in 20% and 40% solutions, respectively.

We take the masses of solutions m (solution 20%) – x, and m (solution 40%) – y, then the mass of the solute in the final solution:

m (r.v.) = m (r-ra 20%) * w1 / 100% + m (r.v. 40%) = m (r-ra 40%) * w2 / 100% = 0.2x + 0,4y = 150

Then the masses of water in the initial solutions are m (in 20%) = m (solution 20%) – m (r.v. 20%) = x – 0.2x = 0.8x and m (in 40%) = m (r-ra 40%) – m (r.v. 40%) = y – 0.4y = 0.6y

The mass of water in the final solution is 0.8x + 0.6y = 350

Let us solve the resulting system of equations

0.2x + 0.4y = 150

0.8x + 0.6y = 350

Let us express x from the first equation: x = (150 – 0.4y) / 0.2 = 750 – 2y, and substitute it into the second equation:

0.8 * (750 – 2y) + 0.6y = 350

600 – 1.6y + 0.6y = 350

600 – 350 = y

y = 250

Then x = 750 – 2 * 250 = 250

m (solution 20%) = 250 g and m (solution 40%) = 250 g