How many grams of a stock solution that is 87.0 percent H2SO4 by mass would be needed to make 200 grams of a 35.5 percent by mass solution? Show all of the work needed to solve this problem.

Initial mass of H2SO4 = final mass of H2SO4 in solution

Initial mass is (87.0%) of target solution (X).
Final mass is (35.5%) of the initial solution (200g).

X = 200g * 35.5/87.0

To solve this problem, we need to use the concept of mass percent and the formula for dilution.

Let's break down the solution into two parts - the stock solution and the final solution.

The stock solution is 87.0% H2SO4, which means it contains 87.0 grams of H2SO4 for every 100 grams of solution.

The final solution is 35.5% by mass, which means it contains 35.5 grams of H2SO4 for every 100 grams of solution.

We need to find the amount of stock solution (in grams) needed to make 200 grams of the final solution.

Let's denote the amount of stock solution needed as 'x' grams.

Now, we can set up a proportion using the mass percent:

(87.0g H2SO4 / 100g stock solution) = (35.5g H2SO4 / 100g final solution)

To solve for 'x':

(87.0g H2SO4 / 100g stock solution) = (35.5g H2SO4 / 100g final solution)
(87.0g / 100g) = (35.5g / 100g)
87.0g = 35.5g * (100g / 100g)
87.0g = 35.5g

Now, we can solve for 'x' using cross-multiplication:

35.5g * (200g / 100g) = x
x = 71.0g

Therefore, we would need 71.0 grams of the stock solution (87.0% H2SO4) to make 200 grams of the final solution (35.5% H2SO4).

chemistry is the most boring subject in the universe. :(