Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml
Answer:
gases
Explanation:
because the solid (steel) is the most dense, the gas (air) is the least dense, and the density of the liquid (water) is in between.
The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.
<h3>What is Keq?</h3>
Keq is the ratio of the concentration of reactant to the concentration of the product.
The balanced equation is
N₂ + 3H₂ = 2NH₃
The equilibrium constant is ![\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5C%3B%20%5BH_2%5D%5E3%7D)
The given concentrations of the compounds have been:
Ammonia = 3 M
Nitrogen = 1 M
Hydrogen = 2 M
Thus, the correct option is D. Keq = 1.5.
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The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.
<h3>What is Ideal law ?</h3>
According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.
PV = nRT.
Where,
- p = pressure
- V = volume (1.75 L = 1.75 x 10⁻³ m³)
- T = absolute temperature
- n = number of moles
- R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
n = PV / RT
State 1 :
- T₁ = (25⁰ C = 25+273 = 298 K)
- p₁ = 225 kPa = 225 x 10³ N/m²
State 2 :
- T₂ = 10 C = 283 K
- p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
Δn = V/R (P₁/T₁ - P₂/T₂ )
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore,
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Hence, The number of moles of gas lost is 0.0213 mol.
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<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.
So the answer is:
8,12,48
</span>
