The answer is 2,507 moles iron.
233,2 g iron contain 4,176 moles.
The mass of 1.23 moles of iron is 58,69 g.
The mass of 4,00 moles of iron is 223,38 g.
FeBr3 (Iron III Bromide) has three moles of bromide for every mole of iron. FeBr2 (Iron II Bromide) has two moles of bromide ion per mole of Iron.
620 g of iron contain 11,1 moles.
111.8g/55.8g= 2.004 moles
18,4 moles of iron contain 1.027,548 g.
0.10 moles iron (55.85 grams/1 mole Fe) = 5.6 grams of iron ==============
0.4965 moles rounded to 4 significant figures
27,9 g iron is equal to 0,5 moles.
The number of iron moles is x(atoms)/6,022140857.10e23.
2,85x1018 atoms of iron are equivalent to 0,73.10-5 moles.
The answer is 0,2686 moles.
7.4 moles iron (55.85 grams/1 mole Fe) = 413.29 grams iron
1.68 grams iron (1 mole Fe/55.85 grams) = 0.030 moles iron ==============
233,2 g iron Feis equivalent to 4,176 moles.
Molar mass of iron is 56g. Given mass of iron= 112g No. of moles = Given mass/Molar Mass => 112g/56g= 2 moles
9,003 x 1023 atoms of iron correspond to 1,495 moles.
The number of moles is 4.
5.0x10^25 * (1 mol / 6.022x10^23 atoms) = 83 moles of iron.
The result is 4,133 moles Fe.
1,25 g of anhydrous iron(III) nitrate = 0,005 moles
divide by Avogadro's number 3.25E24 / 6.022E23 = 5.40 moles
The answer is 1,25 moles iron(III) oxide.