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Answer:

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

Explanation:

The balanced reaction is:

SOCl₂ + H₂O ----> SO₂ + 2 HCl

By stoichiometry of the reaction they react and produce:

  • SOCl₂: 1 mole
  • H₂O: 1 mole
  • SO₂: 1 mole
  • HCl: 2 mole

Being:

  • S: 32 g/mole
  • O: 16 g/mole
  • Cl: 35.45 g/mole
  • H: 1 g/mole

the molar mass of the compounds participating in the reaction is:

  • SOCl₂: 32 g/mole + 16 g/mole + 2*35.45 g/mole= 118.9 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
  • SO₂: 32 g/mole + 2*16 g/mole= 64 g/mole
  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole

Then, by stoichiometry of the reaction, the following amounts of mass react and are produced:

  • SOCl₂: 1 mole* 118.9 g/mole= 118.9 g
  • H₂O: 1 mole* 18 g/mole= 18 g
  • SO₂: 1 mole* 64 g/mole= 64 g
  • HCl: 2 mole* 36.45 g/mole= 72.9 g

Then the following rule of three can be applied: if by stoichiometry of the reaction 118.9 grams of SOCl₂ produce 2 moles of HCl, 0.226 grams of SOCl₂ how many moles of HCl do they produce?

[tex]moles of HCl=\frac{0.226 grams of SOCl_{2}*2 mole of HCl }{118.9 grams of SOCl_{2} }[/tex]

moles of HCl= 3.80*10⁻³

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

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