A. Double-replacement.
2 answers:
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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?
moles of HCl= 3.80*10⁻³
<u><em>3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂</em></u>
Answer:
0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment
Explanation:
The concentration of the solution is given by :
We have:
Concentration of copper (II) nitrate solution =
The volume of solution = 4.12 mL
1 mL= 0.001 L
Moles of copper (II) nitrate in solution = n
Moles of copper (II) nitrate in solution = 0.0099292 mol
1 Mole of copper(II) nitrate has 1 mole of copper then 0.0099292 moles of copper(II) nitrate will have :
Mass of 0.0099292 moles of copper:
This mass of copper present in the solution is the theoretical mass of copper present in the given copper(II) nitrate solution.
0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment