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3 years ago

if 150g of aluminum react with 101g of sulfuric acid what is the maximum amount of hydrogen gas that can be produced

1 answer:

4 0

2Al + 3H2SO4 -> 3H2 + Al2(SO4)3

First, find your limiting reactant - do this with your moles.

Have:
Al - 5.559mol
H2SO4 - 1.030mol
Need:
Al - 0.6867mol
H2SO4 - 8.339mol

Your limiting reactant is sulfuric acid, so we will use that.

Use your mole bridge and the calculated HAVE of sulfuric acid to do this.

1.030mol * (3/3) = 1.030mol of H2

If you want it in liters, multiply that by 22.4dm^3. If you want it in grams, multiply it by 2.016. Hope this helps!

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3 years ago

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A solution of H2SO4 (90.08 g/mol) has 785.6 g of H2SO4 and 359 g of H2O. What is the molal concentration (molality)?

tekilochka [14]

The molality of H₂SO₄ solution is 24.2 m.

Explanation:

We need to find the molality of sulfuric acid.

Mass of sulfuric acid = 785.6 g

Mass of water = 359 g

We have to find the moles of H₂SO₄ by using its mass and molar mass as,

Moles of H₂SO₄ = $$\frac{785.6g}{90.08 g/mol}$

= 8.7 moles

Mass of the solution in kg = $$\frac{359}{1000}$ = 0.359 kg

Molality = $$\frac{moles}{solvent (kg)}$

= $$\frac{8.7 mol }{0.359 kg}\\$

= 24.2 m

So molality of the H₂SO₄ solution is 24.2 m.

8 0

3 years ago

Find a total pressure of the gas mixture that contains forecast is with partial pressure of 5.00 kPa, 4.56 kPa 356 kPa and 678KP

Orlov [11]

Answer: 1044 kPa

Explanation: To find the total pressure of a system, you add up all the partial pressures. 5.00+4.56+356+678=1043.56, but since you need to account for significant figures it is 1044.

In addition, the amount of significant figures is the number with the smallest amount of decimal places. In this case you have no decimal places, since 356 and 678 have no decimal places. Remember that for the numbers 0-4 you round down and 5-9 you round up.

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3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

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murzikaleks [220]

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3 years ago