What mass of sulfur oxide should be added to 500 g of a 20% sulfuric acid solution to increase its mass fraction to 40%.

First, we determine what mass sulfuric acid is in the solution under consideration.
Now let’s create and solve an example.
500 × 20%: 100% = 100 (g) – is sulfuric acid in solution.
Next, we take for x mol the amount of sulfur oxide, SO3, which needs to be added.
The SO3 mass is 80x grams.
Then sulfuric acid H2SO4 is formed, the mass of which is 98x grams.
According to the condition of the problem, a forty percent solution should be obtained.
If we take the entire solution as a unit, then the data 40% will be represented as 0.4.
Now let’s compose and solve the equation.
(100 + 98x) / (500 + 80x) = 0.4.
First, we factor the expression on the left and cancel the resulting fraction.
2 (50 + 49x) / 2 (250 + 40x) = 0.4.
(50 + 49x) / (250 + 40x) = 0.4.
Factor out 2 in the denominator.
(50 + 49x) / 2 (125 + 20x) = 0.4.
Now let’s multiply both sides of the equation by 2, and then cancel and calculate.
2 (50 + 49x) / 2 (125 + 20x) = 0.4 × 2.
(50 + 49x) / (125 + 20x) = 0.4 × 2.
(50 + 49x) / (125 + 20x) = 0.8.
50 + 49x = 0.8 (125 + 20x).
Next, we expand the brackets and calculate.
50 + 49x = 0.8 x 125 + 0.8 x 20x.
50 + 49x = 100 + 16x.
49x – 16x = 100 – 50.
33x = 50.
x = 50: 33.
x = 1 17/33.
Now let’s represent the given value of x as a decimal fraction.
x = 1 17/33 = 1, (51) = 1.52.
Next, we find the mass of SO3.
80x = 80 × 1.52 = 121.6 (d).
Answer: 121.6 grams.