Respuesta :

joseaaronlara

Answer:

The answer to your question is 588 g of H₂SO₄

Explanation:

Data

mass of H₂SO₄ = ? x

volume = 500 ml = 0.5 liters

concentration = molarity = 12 M

Formula

Molarity = [tex]\frac{moles}{volume}[/tex]

Process

1.- Solve for moles

   moles = Molarity x volume

   moles = 12 x 0.5

   moles = 6

2.- Calculate the molecular mass of H₂SO₄

Molecular mass = (2 x 1) + (32 x 1) + (16 x 4) = 98 g

3.- Use proportions to find the moles of sulfuric acid in the solution

                              1 mol of H₂SO₄   ---------------  98 g

                              6 moles of H₂SO₄ ------------   x

                                 x = (6 x 98) / 1

                                x = 588 g

ajibodeet

Answer:

2352grams

Explanation: To calculate the number of gram of H₂SO₄ present in 500-milliliter of solution 12.0 M (moles) of H₂SO₄,

•We have to calculate the concentration of H2SO4 solution

•Calcualate the mass of H2SO4 i n the solution

Since, Number of moles = Concentration of solution × volume of solution

Concentration of H2SO4= number of moles of H2SO4 ÷ volume of H2SO4 solution

No of moles= 12 moles

Volume = 500mL = 500/1000L =0.5L

Concentration of H2SO4 = 12moles÷ 0.5L

Concentration of H2SO4= 24 mol/L or 24mol/dm^3

To calculate mass of H2SO4 in gram

•Concentration of H2SO4= mass of H2SO4 ÷ molar mass of H2SO4

Molar mass of H2SO4= (2×1) + 32 + (4×16)

Molar mass of H2SO4= 98g/mol

•Molar mass of H2SO4 in g/dm^3= Concentration of H2SO4 solution in mol/dm^3 × Molar mass of H2SO4 g/mol

Mass of H2SO4= 24mol/dm^3 × 98g/mol

Mass of H2SO4 present= 2352g/dm^3

Therefore 2352grams of H₂SO₄ are in a 500-milliliter of solution 12.0 moles of H₂SO₄ solution.

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