Respuesta :
the balanced equation for the above reaction is
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
The concentration of H_2SO_4 is 0.125 M. Molarity of any compound is the concentration of that compound.
What is concentration?
Concentration is the quantity of any compound in a particular space.
The reaction is [tex]\rm 2NaOH + H_2SO_4--- > Na_2SO_4 + 2H_2O[/tex]
The ratio of NaOH to H₂SO₄ is 2:1
Calculate the number of moles of NaOH
[tex]\rm \dfrac{0.5000 M }{1000 mL/L} /\times 21.17 mL = 0.010585 mol[/tex]
There are 2 moles of NaOH, thus to find the moles of H₂SO₄, we will divide the mole of NaOH by 2
[tex]\dfrac{\dfrac{0.010585}{2}}{ 42.35\;ml} \times 1000 ml = 0.125 M.[/tex]
Thus, the molarity of hydrochloric acid is 0.125 M
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