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Pani-rosa [81]
3 years ago
8

0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +

3H2SO4(aq) -------->Al2(SO4)3(aq) + 6H2O(l) How many moles of H2O can form under these conditions?
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
5 0

Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

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If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr
Fynjy0 [20]
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
7 0
4 years ago
Microscale glassware kits include connectors to fit the smaller glassware. The black connectors should be used when _________ ar
spayn [35]

Answer:

high boiling aromatics

with the same ground glass joint.

Explanation:

The black connectors are more resistant to the reactant than the white connectors. This implies that the black connectors should be used when high boiling aromatics are present, Else, the tan connections can be used.

Thus, the connectors can combine different pieces of glassware or components <u>with the same ground glass joint.</u>

<u />

6 0
3 years ago
A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that fo
amid [387]

Explanation:

Ksp of NiCO3 = 1.4 x 10^-7

Ksp of CuCO3 = 2.5 x 10^-10

Ionic equations:

NiCO3 --> Ni2+ + CO3^2-

CuCO3 --> Cu2+ + CO3^2-

[Cu2+][CO3^2-]/[Ni2+][CO3^2-]

= (2.5* 10^-10)/(1.4* 10^-7)

= 0.00179.

[Cu2+]/[Ni2+]

= 0.00179

= 0.00179*[Ni2+]

If all of Cu2+ is precipitated before Na2CO3 is added.

= 0.00179 * (0.25)

The amount of Cu2+ not precipitated = 0.000448 M

The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100

= 0.000448/0.25 * 100

= 0.18%

Therefore, percentage precipitated = 100 - 0.18

= 99.8%

The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.

8 0
3 years ago
Which of the following groups of elements are likely to have substances that have both metal and non-metallic properties?
Tom [10]

Answer : Group 14 elements are likely to have substances that have both metal and non-metallic properties.

Explanation :

Metalloid : It is a type of element whose properties lies in between the metals and non-metals.

There are commonly six metalloids : Boron(group 13), Silicon & Germanium(group 14), Arsenic & Antimony(group 15), Tellurium(group 16).

Properties of Metalloids :

  • Metalloids look like metals but behave largely like non-metals.
  • Metalloids are solid, brittle and lustrous.
  • Metalloids has ability to conduct electricity.

Therefore, Group 14 elements are likely to have substances that have both metal and non-metallic properties.


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