How many milliliters of 0.424 M FeCl2 solution are needed to react completely wi

Question

How many milliliters of 0.424 M FeCl2 solution are needed to react completely with 41.1 mL of 0.275 M NaOH solution? How many grams of Fe(OH)2 will be formed? The reaction is

2NaOH(aq) + FeCl2(aq) ? Fe(OH)2(s) + 2NaCl(aq)



(a) mL
(b) g

Explanation / Answer

2NaOH(aq) + FeCl2(aq) ----> Fe(OH)2(s) + 2NaCl(aq) No . of moles of NaOH , n = Molarity * Volume in L = 0.275 M * 0.0411 L = 0.0113 moles According to the Equation , 2 moles of NaOH reacts with 1 mole of FeCl2 0.0113 moles of NaOH reacts with 0.0113/2 moles of FeCl2 = 0.00565 moles of FeCl2 Volume of FeCl2 required , V in L = no . of moles / Molarity = 0.00565 mol / 0.424 M = 0.0133 L = 13.3 mL ---------------------------------------------------- According to the Equation , 2 moles of NaOH produces 1 mole of Fe(OH)2 0.0113 moles of NaOH produces 0.0113 / 2 = 0.00565 moles Molar mass of Fe(OH)2 = 56 +2( 16+1) = 90 g / mol mass of Fe(OH)2 , m = no . of moles * Molar mass = 0.00565 * 90 = 0.5085 g

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