I'm assuming that by "miles" you mean moles.
If O2 is the excess reactant, that means Fe is the limiting reactant. That means that the amount of product being formed depends on the amount of Fe reactant present. To calculate the moles of Fe2O3 formed, start with the given 6.4 moles of Fe and use the mole to mole ratio given by the reaction as shown below:
6.4 mol Fe x

=
3.2 mol Fe2O3
Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
Answer:
Mass of solute = 0.0036 g
Explanation:
Given data:
Concentration of Cl⁻ = 15.0 ppm
Volume of water = 240 mL
Mass of Cl⁻ present = ?
Solution:
1 mL = 1 g
240 mL = 240 g
Formula:
ppm = mass of solute / mass of sample ×1,000,000
by putting values,
15.0 ppm = (mass of solute / 240 g) ×1,000,000
Mass of solute = 15.0 ppm × 240 g / 1,000,000
Mass of solute = 0.0036 g