Enter your answer in the provided box. Before arc welding was developed, a displacement reaction involving aluminum and iron(III) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 1.64 kg of aluminum reacts with 20.2 mol of iron(III) oxide.

grams of iron = 2262.4  g

Explanation:

The balanced chemical equation should be represented first.  

2Al + Fe2O3 → Al2O3 + 2Fe

find the limiting reactants by converting to moles.

covert 1.64 kg to grams = 1640 g

moles = mass/molar mass = 1640/27 = 60.7407407407  moles

60.7407407407 × 1 mol-rxn /2 = 30.37  mol-rxn

The limiting reactant is Fe2O3 Therefore it will determine the yield of Iron

2Al + Fe2O3 → Al2O3 + 2Fe

molar mass of Fe2O3 = 56 × 2 + 16 × 3 =  112 + 48 = 160 g

atomic mass of iron = 56

mass of iron in the reaction = 56 × 2 = 112 g

moles = mass/molar mass

mass = moles × molar mass

mass = 20.2 × 160 = 3232  g

if 160 g of Fe2O3  give 112 g of iron

3232 g of Fe2O3 will give ? grams of iron

grams of iron = 3232 × 112/ 160

grams of iron = 361984 /160

grams of iron = 2262.4  g

i believe it is a.) released into

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%