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
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Ответ:
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) =
5.56g of FeCl2 can be producedОтвет:
A
Explanation: