The problem provides a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. Therefore, to determine the quantity of red blood cells in your sample, we must first convert Liters to cm3 using the conversion factor--> 1 mL= 1 cm3
I have shared how to resolve this issue.
84.34 grams of iron (III) chloride is the maximum produced since iron is the limiting reagent, and chlorine gas is in excess.
Explanation:
Balanced equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA PROVIDED:
iron = atoms
mass of chlorine = 67.2 liters
mass of FeCl3 =?
The number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 mol of iron
moles of chlorine gas
number of moles = 
Substituting the values into the equation:
n =
(molar mass of chlorine gas = 70.96 g/mol)
= 947.01 moles
As iron is the limiting reagent therefore
2 moles of Fe lead to 2 moles of FeCl3
0.52 moles of Fe will yield
= 
0.52 moles of FeCl3 is produced.
To express this in grams:
mass = n x molar mass
= 0.52 x 162.2 (molar mass of FeCl3 is 162.2g/mol)
= 84.34 grams
We need to calculate the volume of Gold, assuming its mass matches that of copper.
Given information:
Density of Copper = 8.96 g/ml.
Volume of Copper = 141 ml.
Mass of Gold = Mass of Copper.
Density of Gold = 19.3 g/ml.
To find copper's mass, we use the density equation:
Density = mass/volume.
To find mass of copper:
Mass of copper = Density of Copper * Volume of Copper.
Mass of copper = 8.96 g/ml * 141 ml = 1263.36 g.
Thus,
Mass of gold = Mass of copper = 1263.36 g.
Now, using the density formula for gold to get its volume:
Volume of gold = Mass of gold / Density of gold.
Volume of gold = 1263.36 g / 19.3 g/ml = 65.46 mL.
Consequently, the volume of gold required to match the mass of copper is 65.46 mL.