The answer would be option 4
The answer is A. The m and n are the coefficients obtained from the equation. And important thing is that the equation need to be balanced before using its coefficients.
The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>
Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

Answer:
56160grams
Explanation:
First, we need to convert the number of molecules of magnesium chloride (MgCl2) into moles by dividing by Avagadro's number (6.02 × 10^23 molecules)
n = nA ÷ 6.02 × 10^23
n = 14.07 × 10^26 ÷ 6.02 × 10^23
n = 14.07/6.02 × 10^(26-23)
n = 2.34 × 10^3 moles of MgCl2
The balanced reaction given in the question is as follows:
Mg + 2HCl → MgCl2 + H2
If 1 mole of Mg produced 1 mole of MgCl2
Then, 2.34 × 10^3 moles of Mg will also produce 2.34 × 10^3 moles of MgCl2.
Using mole = mass ÷ molar mass (MM)
Molar mass of Mg = 24g/mol
mass = mole × MM
mass = 2.34 × 10^3 × 24
mass = 56.16 × 10^3
mass = 56160grams.