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Lemur [1.5K]
4 years ago
12

In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure

to make H₂SO₄, which is the limiting reagent?
Chemistry
1 answer:
nikitadnepr [17]4 years ago
6 0

Answer:

The limiting reagent is the O₂

Explanation:

We can think, this reaction

2O₂(g) + H₂(g) + S(s)  →  H₂SO₄

Mole of each = Mass / molar mass

6 g / 32 g/m = 0.187 mole O₂

4g / 2 g/m = 2 mole H₂

5g / 32.06 g/m = 0.156 mole S

Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

0.187 mole of O₂, I need (the half)

0.093 mole of H₂ and 0.093 mole of S

For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

2 mole of H₂, I need (the double of O₂ and the same for S)

4 mole of O₂ ; 2 mole of S

For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

0.156 mole I need the same amount for H₂ and the double for O₂

0.156 mole of H₂ and 0.312 mole of O₂

In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

4 0
4 years ago
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think i
nlexa [21]

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

5 0
3 years ago
2. List and elaborate on at least two limitations of the 24-hour recall as a
Pavlova-9 [17]

Answer:

The correct answer is - may not be typical, and participant burden.

Explanation:

The 24-hour recall is nothing but a retrospective method of diet assessment. In this method, an individual is interviewed about his or her diet consumption during the last 24 hours.

The disadvantages or limitations of this method include the inability of a single day's intake to describe the typical diet, multiple recalls to intake, cost and administration time; participant burden, have to recall to reliably estimate usual intake.

5 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
What happens as a result of an increase in the intensity of a sound wave?
svet-max [94.6K]
The result is the sound will be louder. This is due to the high amount of energy in the sounds.
3 0
4 years ago
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