Question

Визначити об'єм 0,5М розчину сульфатної кислоти, необхідного для повного розчинення 12 г магнію

Answer 1

Ответ:

1 литр

Объяснение:

n(Mg)=12/24=0.5 моль

Mg + H2SO4 = MgSO4 + H2O

n(Mg)=n(H2SO4)

n(H2SO4)=CmV

V(H2SO4)=0.5/0.5=1 литр

Answer 2

Calculating the Volume of 0.5M Sulfuric Acid Solution Required to Completely Dissolve 12g of Magnesium

To determine the volume of 0.5M sulfuric acid solution needed to completely dissolve 12g of magnesium, we can use the chemical equation for the reaction between sulfuric acid and magnesium. The balanced chemical equation for the reaction is:

H2SO4 + Mg → MgSO4 + H2

From this equation, we can see that 1 mole of sulfuric acid reacts with 1 mole of magnesium. The molar mass of sulfuric acid (H2SO4) is 98.08 g/mol, and the molar mass of magnesium (Mg) is 24.305 g/mol.

Using this information, we can calculate the amount of sulfuric acid needed to react with 12g of magnesium.

Calculation:

1. Calculate the moles of magnesium: Moles of Mg = Mass of Mg / Molar mass of Mg Moles of Mg = 12g / 24.305 g/mol Moles of Mg = 0.493 moles

2. Since the reaction is 1:1 between sulfuric acid and magnesium, we need the same number of moles of sulfuric acid as moles of magnesium to completely react.

3. Now, we can calculate the volume of 0.5M sulfuric acid solution needed using the formula: Volume (in liters) = Moles / Molarity Volume = 0.493 moles / 0.5 mol/L Volume ≈ 0.986 L

Therefore, approximately 0.986 liters of 0.5M sulfuric acid solution are required to completely dissolve 12g of magnesium.