Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
The radius of the anion is 7.413 nm
<h3>How to calculate the force of attraction between charges</h3>
The force of attraction (F) is given by the formula:
- F = (1/4π∈r²)(Zc*e)(Za*e)
where:
∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m
Zc = charge on the cation = +2
Zc = charge on the anion = -2
e = charge on an electron = 1.602 * 10⁻¹⁹ C
r = interionic distance
r = rc + ra
where rc and ra are the radius of the cation and anion respectively
F = 1.64 * 10⁻⁸ N
Therefore based on the equation of force of attraction:
1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²
r² = 5.63 * 10⁻¹⁷
r = 7.50 nm
Since r = rc + ra
where rc = 0.087 nm
thus, ra = r - rc = 7.50 - 0.087
ra = 7.413 nm
Therefore, the radius of the anion is 7.413 nm
Learn more about ionic radius at: brainly.com/question/2279609
4 degrees Celsius I think
Answer:
0.1739
Explanation:
0.800 mol of N2
0.200 mol of H2
0.150 mol of CH4
Total moles of the mixture = 0.8 + 0.2 + 0.150 = 1.150 mol
Mole fraction of H2 = Number of moles of H2 / Total moles
Mole Fraction = 0.2 / 1.150 = 0.1739
