If 2.45 g of iron are placed in 1,5 L of 0.25M HCl, how many grams of FeCl2 are obtained? Identify the limiting and excess reactants in this single replacement reaction. Fe + 2HCl = FeCl2 + H2

5.56g of FeCl2 can be produced

Explanation:

To solve this question we must find the moles of each reactant. With the moles and the chemical equation we can find limiting reactant. With limiting reactant we can find the moles of FeCl2 and its mass as follows:

Moles HCl:

1.5L * (0.25mol / L) = 0.375 moles HCl

Moles Fe -Molar mass: 55.845g/mol-

2.45g * (1mol / 55.845g) = 0.0439 moles Fe

For a complete reaction of 0.375 moles HCl are needed:

0.375 moles HCl * (1mol Fe / 2mol HCl) = 0.1875 moles Fe

As there are just 0.0439 moles Fe, Fe is limiting reactant

1mol of Fe produce 1 mole of FeCl2, 0.0439 moles Fe produce 0.0439 moles of FeCl2. The mass is:

Mass FeCl2 -Molar mass: 126.751g/mol:

0.0439 moles Fe * (126.751g / mol) =

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"