4.80 dm^3 of chlorine gas was reacted with NaOH solution.
a) how many moles of Cl2 reacted
b) what mass of NaOCl was formed?
c) if the concentration of the NaOH was 2.00 mol/dm^3. What volume of sodium hydroxide solution was required?
d) write an ionic equation for this reaction
Chemistry moles help please?
2016-07-12 5:23 pm
回答 (2)
2016-07-12 5:37 pm
(a)
At RTP, molar volume of Cl₂ gas = 24 dm³/mol
No. of moles of Cl₂ reacted = (4.80 dm³) / (24 dm³/mol) = 0.200 mol
(b)
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O
OR: Mole ratio Cl₂ : NaOCl = 1 : 1
No. of moles of Cl₂ reacted = 0.200 mol
No. of moles of NaOCl reacted = 0.200 mol
Molar mass of NaOCl = (22.99 + 16.00 + 35.45) g/mol = 74.44 g/mol
Mass of NaOCl formed = (74.44 g/mol) × (0.200 mol) = 14.9 g
(c)
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O
OR: Mole ratio Cl₂ : NaOH = 1 : 1
No. of moles of Cl₂ reacted = 0.200 mol
No. of moles of NaOH required = 0.200 mol
Volume of NaOH required = (0.200 mol) / (2.00 mol/dm³) = 0.100 dm³ = 100 cm³
(d)
Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O
At RTP, molar volume of Cl₂ gas = 24 dm³/mol
No. of moles of Cl₂ reacted = (4.80 dm³) / (24 dm³/mol) = 0.200 mol
(b)
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O
OR: Mole ratio Cl₂ : NaOCl = 1 : 1
No. of moles of Cl₂ reacted = 0.200 mol
No. of moles of NaOCl reacted = 0.200 mol
Molar mass of NaOCl = (22.99 + 16.00 + 35.45) g/mol = 74.44 g/mol
Mass of NaOCl formed = (74.44 g/mol) × (0.200 mol) = 14.9 g
(c)
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O
OR: Mole ratio Cl₂ : NaOH = 1 : 1
No. of moles of Cl₂ reacted = 0.200 mol
No. of moles of NaOH required = 0.200 mol
Volume of NaOH required = (0.200 mol) / (2.00 mol/dm³) = 0.100 dm³ = 100 cm³
(d)
Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O
2016-07-12 5:51 pm
1st ... you need a balanced equation
NaOH + Cl2 --> NaOCl + HCl(g) ???
or 2NaOH + Cl2 --> NaOCl + NaCl + H2O <<< probably
.... I am using the 2nd equation
a. assuming STP for the gas ...
80 dm^3 X (1mol / 22.4 dm^3) = 3.57 mole Cl2
b. 3.57 mole X [1mole NaOCl / 1 mole Cl2] X [ 74.5g / mole ]
... use a calculator
c. 3.57 mole X [2 mol NaOH / 1 mole Cl2] X [ 1 dm^3 / 2.00mol]
... use a calculator ... gives volume in dm^3 ..
multiply be 1000 if you want volume in cm^3
d. 2Na(+) + 2OH(-) --> 2Na(+) + OCl(-) + Cl(-) + H2O
NaOH + Cl2 --> NaOCl + HCl(g) ???
or 2NaOH + Cl2 --> NaOCl + NaCl + H2O <<< probably
.... I am using the 2nd equation
a. assuming STP for the gas ...
80 dm^3 X (1mol / 22.4 dm^3) = 3.57 mole Cl2
b. 3.57 mole X [1mole NaOCl / 1 mole Cl2] X [ 74.5g / mole ]
... use a calculator
c. 3.57 mole X [2 mol NaOH / 1 mole Cl2] X [ 1 dm^3 / 2.00mol]
... use a calculator ... gives volume in dm^3 ..
multiply be 1000 if you want volume in cm^3
d. 2Na(+) + 2OH(-) --> 2Na(+) + OCl(-) + Cl(-) + H2O
收錄日期: 2021-04-20 16:30:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160712092304AAm9CSD