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.
Answer:
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 261.34 233.39
Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃
m/g: 75.00
1. Moles of Ba(NO₃)₂
2. Moles of BaSO₄
The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂
3. Mass of BaSO₄
Answer:
4 moles
Explanation:
A mole is equal to 6.02214076 × 1023 of any chemical unit (atoms, molecules, ions)
To find number of moles in 89 litres of water vapor use the following formula:
1 mole = 22.4 L
That is 1 L = 
mole
Volume of water vapor = 89 litres
Therefore,
Number of moles in 89 litres = 
≈ 4 moles
Answer:
pH = 0.35
Explanation:
For a strong acid, all of the acid dissociates into H3O+, and pH = -log[H3O+], where [H3O+] = [HClO3] = 0.45 M.
<span>mv^2 is a conserved quantity
so v goes as sqrt(1/m)
heavier is slower
64/4 = 16
sqrt(16) = 4
B) and D) are both correct strictly speaking - lean towards submitting D) as answer
so i would say D) </span>
