# Question aae95

May 1, 2015

The copper (II) sulfate will be the excess reactant by 6 g.

$F {e}_{\left(s\right)} + C u S {O}_{4 \left(a q\right)} \to F e S {O}_{4 \left(a q\right)} + C {u}_{\left(s\right)}$

Notice that you have a $1 : 1$ mole ratio between iron and copper (II) sulfate, which implies that the number of moles of that reacted will be equal for both reactants.

Whichever reactant will have more moles present, that will be your excess reactant. Determine how many moles of each reactant are present by using their respective molar mass

2cancel("g") * "1 mole Fe"/(55.845cancel("g")) = "0.0358 moles Fe"

12cancel("g") * ("1 mole "CuSO_4)/(159.609cancel("g")) = "0.0752 moles " $C u S {O}_{4}$

Since you have more moles of copper (II) sulfate than you have of iron, copper (II) sulfate will be the excess reactant (iron will be the limiting reagent).

The reaction will consume all of the iron and leave you with

n_(CuSO_4"excess") = 0.0752 - 0.0358 = "0.0394 moles" $C u S {O}_{4}$

This is equivalent to having an excess of

0.0394cancel("moles"CuSO_4) * "159.609 g"/(1cancel("mole" CuSO_4)) = "6.29 g"#

Rounded to one sig fig, the number of sig figs given for the mass of iron, the answer will be

${m}_{C u S {O}_{4} \text{excess") = color(green)("6 g}}$