Answer:
.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?
moles of HCl= 3.80*10⁻³
3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂
