Calculate the number of moles of iron(III)sulfide that forms when 62.0 mL of 0.135 M iron(III)chloride reacts with 45.0 mL of 0.285 M calcium sulfide.
a) 8.56 x 10−3 mol
b) 4.19 x 10−3 mol
c) 1.25 x 10−2 mol
d) 4.28 x 10−3 mol
e) 8.38 x 10−3 mol
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