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MrRa [10]
3 years ago
13

How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?

Chemistry
1 answer:
vova2212 [387]3 years ago
7 0

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

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notka56 [123]

Answer:

pH=4.56

Explanation:

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In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

[base]=0.000781mol/0.025L=0.0312M

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

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6 0
2 years ago
7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.
Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

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3 years ago
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