Question

Высчитайте объём водорода который может быть получен при взаимодействии 400г. 49-% серной кислоты с магнием

Answer 1

Mg+H2SO4=MgSO4+H2
             1-------------------1
400*0.49=196 г H2SO4
n(H2SO4)=196:98=2 моль=n(H2)
V=22.4*2=44.8 л

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when 400g of 49% sulfuric acid reacts with magnesium, we can use the balanced chemical equation for the reaction and the concept of molar volume at standard temperature and pressure (STP).

The balanced chemical equation for the reaction between sulfuric acid and magnesium is:

H2SO4 + Mg → MgSO4 + H2

First, we need to calculate the number of moles of sulfuric acid and magnesium involved in the reaction.

Calculating Moles of Sulfuric Acid

The molar mass of sulfuric acid (H2SO4) is 98.08 g/mol. Since the sulfuric acid is 49% by mass, the mass of pure sulfuric acid in 400g of the solution is: \[400g \times \frac{49}{100} = 196g\]

The number of moles of sulfuric acid can be calculated using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] \[ \text{moles} = \frac{196g}{98.08g/mol} = 2 moles \]

Calculating Moles of Magnesium

The molar mass of magnesium (Mg) is 24.305 g/mol. Since the reaction involves stoichiometry, we only need one mole of magnesium to react with one mole of sulfuric acid.

Now, we can calculate the volume of hydrogen gas produced using the concept of molar volume at STP, which is 22.4 liters per mole.

Calculating Volume of Hydrogen Gas

Since the reaction produces one mole of hydrogen gas for every mole of sulfuric acid, the volume of hydrogen gas produced can be calculated as: \[ \text{Volume} = \text{moles} \times \text{molar volume at STP} \] \[ \text{Volume} = 2 \times 22.4 = 44.8 \text{ liters} \]

Therefore, the volume of hydrogen gas that can be obtained from the reaction of 400g of 49% sulfuric acid with magnesium is 44.8 liters.