For this you need the Atomic Mass of Fe. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.
82.5 moles Fe / (55.9 grams) = 1.48 moles Fe
The chemical formula of iron(II) chloride is FeCl2.
Molecular mass of FeCl2 = 55.8 + 2(35.5) = 126.8
Amount of FeCl2 in a 85.2g sample = 85.2/126.8 = 0.672mol
The answer is 125,65 g.
FeCl3 ? 1.1 grams FeCl3 (1 mole FeCl3/162.2 grams) = 0.0068 moles of FeCl3 ------------------------------------
Not many! 5.0 grams CaBr2 (1 mole/199.88 grams CaBr2) = 0.025 moles CaBr2
Multiple the moles of K2SO4 by the molecular weight of 174.2592 grams. That should equal 210.85 grams.
For this you need the atomic (molecular) mass of CO2. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. CO2=44.0 grams454 grams CO2 / (44.0 grams) = 10.3 moles CO2
7.4 moles iron (55.85 grams/1 mole Fe) = 413.29 grams iron
738 grams iron are equivalent to:- 12,626 moles- 76.10e23 atoms
23 grams iron ( 1mole Fe/55.85 grams) = 0.41 moles of iron
Let's see. 3 moles gold (197.0 grams/1 mole Au) = 591 grams of gold ----------------------------- 10 mole iron (55.85 grams/1 mole Fe) = 558.5 grams of iron ----------------------------- So, 3 moles of gold has more mass that 10 moles of iron. (heavier)
To calculate the number of moles from grams, you must divide by the substance's molar mass
1,25 g of anhydrous iron(III) nitrate = 0,005 moles
The mass of 4 moles of iron is 224 grams
388.grams Fe divided by mass of Fe 388.2g/55.85= 6.95moles of Fe
This cannot be answered because it is essential to know what element you're dealing with here. 8.2 grams of iron will contain considerably less atoms than 8.2 of hydrogen. Therefore, the number of moles in 8.2 grams of iron will differ from the number of moles in 8.2 grams of hydrogen.
75.89 grams of FeO is equal to 1.06 moles of iron II oxide. In decomposition, there would be 1.06 moles of iron recovered, or 58.99 grams.
The molecular mass of iron(III) chloride is 55.8 + 3(35.5) = 162.3 Amount of iron (III) chloride in a 80.5g pure sample = 80.5/162.3 = 0.496mol
825 grams = 825,000 mg. 1 gr. = 1000 mg.