Calculate the mass of a 15% sulfuric acid solution that reacted with magnesium if a salt of 240 grams was formed.

Data: ωН2SO4 – mass fraction in the taken sulfuric acid solution (ωН2SO4 = 15% = 0.15); mМgSO4 is the mass of the formed salt (mМgSO4 = 240 g).

Const: MМgSO4 – molar mass of magnesium sulfate (MМgSO4 = 120.37 / mol ≈ 120 g / mol); MН2SO4 – molar mass of sulfuric acid (MН2SO4 = 98.078 g / mol ≈ 98 g / mol).

1) Amount of salt formed (magnesium sulfate): νМgSO4 = mМgSO4 / MМgSO4 = 240/120 = 2 mol.

2) The reaction level: Mg (magnesium) + Н2SO4 (taken solution, sulfuric acid) = МgSO4 (precipitate, magnesium sulfate) ↓ + Н2 (hydrogen) ↑.

3) Amount of sulfuric acid thing: νМgSO4 / νН2SO4 = 1/1 and νН2SO4 = 2 mol.

5) Mass of sulfuric acid: mН2SO4 = MН2SO4 * νМgSO4 = 98 * 2 = 196 g.

4) The required mass of the solution: mr-pa = mН2SO4 / ωН2SO4 = 196 / 0.15 = 1306, (6) g.

Answer: The mass of the sulfuric acid solution taken was 1306, (6) g.



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