The balanced chemical reaction is written as:
Sb2S3 + 6HCl = 6SbCl<span>3 + 3H2S
We are given the amount of </span><span>antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:
2.85 g Sb2S3 ( 1 mol / </span><span>339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) (</span> 228.13 g / mol ) = 11.48 g SbCl3
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
Answer:
The factor of increasing reaction rate is 1,85x10¹².
Explanation:
Using arrhenius formula:
Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K
Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:
k = 1,25x10⁻²²
When activation energy is 55kJ/mol:
k = 2,31x10⁻¹⁰
Thus, the factor of increasing reaction rate is:
2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>
<em></em>
I hope it helps!
Answer:
HF is the limiting reactant
Explanation:
The balanced equation for the reaction is given below:
SiO₂ + 4HF —> SiF₄ + 2H₂O
From the balanced equation above,
1 mole of SiO₂ reacted with 4 moles of HF.
Finally, we shall determine the limiting reactant. This can be obtained as illustrated below:
From the balanced equation above,
1 mole of SiO₂ reacted with 4 moles of HF.
Therefore, 7.5 moles of SiO₂ will react with = 7.5 × 4 = 30 moles of HF.
From the calculation made above, we can see clearly that it will take a higher amount (i.e 30 moles) of HF than what was given from the question (i.e 5 moles) to react completely with 7.5 moles of SiO₂.
Therefore, HF is the limiting reactant and SiO₂ is the excess reactant.
