25.0 mL of a 1.00 M Fe(No3)2 solution was reacted with 25.0 mL of a 0.700M...

25.0 mL of a 1.00 M Fe(No3)2 solution was reacted with 25.0 mL of a 0.700M solution of K3PO4 to produce the solid Fe3(PO4)2 by the balanced chemical equation

3 Fe(NO3)2(aq)+ 2 K3PO4(aq)--->Fe3(PO4)2(s)+ 6 KNO3(aq)

What is the limiting reagent?

How many grams of Fe(PO4)2 will be produced?

Solutions

Expert Solution

Number of moles of Fe(NO3)2 = molarity * volume of solution in L

Number of moles of Fe(NO3)2 = 1.00 * 0.025 = 0.025 mole

Number of moles of K3PO4 = molarity * volume of solution in L

Number of moles of K3PO4 = 0.700 * 0.025 = 0.0175 mole

From the balanced equation we can say that

3 mole of Fe(NO3)2 requires 2 mole of K3PO4 so

0.025 mole of Fe(NO3)2 will require

= 0.025 mole of Fe(NO3)2 *(2 mole of K3PO4 / 3 mole of Fe(NO3)2)

= 0.0167 mole of K3PO4

But we have 0.0175 mole of K3PO4 so K3PO4 is an excess reactant and Fe(NO3)2 is a limiting reactant

From the balanced equation we can say that

3 mole of Fe(NO3)2 produces 1 mole of Fe3(PO4)2 so

0.025 mole of Fe(NO3)2 will produce

= 0.025 mole of Fe(NO3)2 *(1 mole of Fe3(PO4)2 / 3 mole of Fe(NO3)2)

= 0.00833 mole of Fe3(PO4)2

mass of 1 mole of Fe3(PO4)2 = 357.48 g so

the mass of 0.00833 mole of Fe3(PO4)2 = 2.98 g

Therefore, the mass of Fe3(PO4)2 produced would be 2.98 g

queen_honey_blossom answered 1 year ago

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