Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
1°/ . 2 Al + 6 HCl → 2 AlCl3 + 3 H2
<span>k1 = n(Al) / 2 = 4,5 / 2 = 2,25 </span>
<span>k2 = n(HCl) / 6 = 11,5 / 6= 1,92 </span>
<span>k2 < k1 ==> HCl is the limiting reactant </span>
<span>6 mol of HCl ---> 2 mol of H2 </span>
<span>11,5 mol of HCl ---> 3,83 mol of H2 </span>
Answer:
1.02mol
Explanation:
Using the general gas equation below;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information provided in this question,
P = 2.0 atm
V = 11.4L
T = 273K
n = ?
Using PV = nRT
n = PV/RT
n = 2 × 11.4/ 0.0821 × 273
n = 22.8/22.41
n = 1.017
n = 1.02mol
Answer:
3.33 L
Explanation:
We can solve this problem by using the equation:
Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:
We input the data:
- 0.25 M * 100 L = 7.5 M * V₂
Thus the answer is 3.33 liters.
